Chapter 3 – Data Representation

 

The focus of this chapter is the representation of data in a digital computer.  We begin with a
review of several number systems (decimal, binary, octal, and hexadecimal) and a discussion
of methods for conversion between the systems.  The two most important methods are
conversion from decimal to binary and binary to decimal.  The conversions between binary
and each of octal and hexadecimal are quite simple.  Other conversions, such as conversion
to and from base–5 are of interest only to teachers with a bad sense of humor.

 

After discussion of conversion between bases, we discuss the methods used to store integers
in a digital computer: one’s complement and two’s complement arithmetic.  This includes a
characterization of the range of integers that can be stored given the number of bits allocated
to store an integer.  The most common integer storage formats are 16 and 32 bits. 

 

The next topic for this chapter is the storage of real (floating point) numbers.  This discussion
will focus on the standard put forward by the Institute of Electrical and Electronic Engineers,
the IEEE Standard 754 for floating point numbers.  The chapter closes with a discussion of
codes for storing characters: ASCII, EBCDIC, and Unicode.

 

General Remark on Data Storage in a Computer

Data (numeric, character, and more complex structures) are all stored in the computer and
on disk in binary form.  Each common storage medium supports only two values, either two
distinct voltages or two distinct magnetic states, or some other two–state system.  The most
common way to denote these two states is based on the binary number system.

Consider what happens when a data value is read from computer memory.  Each value is
encoded as a number of bits (binary values: 0 or 1) and stored in the same number of memory
cells, which we shall discuss later.  When a value is read, each cell returns a voltage to the
reading circuitry: either a positive voltage (say 1.5 volts), interpreted as a logic 1, or a zero
voltage, interpreted as a logic 0.  It is important to note that the actual value of the positive
voltage is so dependent on technology that we can say little about its actual value.

A stored–program computer (that is, almost any computer or digital device now in use) stores
both data and program instructions in memory.  Each item stored in memory is best viewed
as nothing more than a collection of binary bits.  What that item signifies depends on how
that item is used by the computer.  An item may be used as an instruction, as an address of
another item, or as data in one of many formats.

When we study some modern memory architectures, we shall discover several mechanisms
that make it less likely that an item of one type will be used in an incorrect context.  While
these mechanisms are widespread, they are additions to the basic computer design and not
fundamental to it.  As an example, an unfortunate programming error by the author of these
notes caused the stored representation of the real number 2.5 to be executed as if it were a
valid instruction.  The problem with that was that it executed properly and redirected the
program to another area, where the program abruptly ceased execution.

 


Number Systems

There are four number systems of possible interest to the computer programmer: decimal,
binary, octal, and hexadecimal.  Each system is characterized by its base or radix, always
given in decimal, and the set of permissible digits.  For small sets, such as each of these,
we specify the set by listing all of its members; for example the set of ten decimal digits.

Sets specified by listing all items belonging to the set often present the items in some
sort of order, but that is not necessary, as sets are by definition unordered collections.

Note that the hexadecimal numbering system calls for more than ten digits, so we use the
first six letters of the alphabet.  It is preferable to use upper case letters for these “digits”.

Decimal              Base = 10
                           Digit Set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Binary                Base = 2
                           Digit Set = {0, 1}

Octal                  Base = 8 = 23
                           Digit Set = {0, 1, 2, 3, 4, 5, 6, 7}

Hexadecimal      Base = 16 = 24
                           Digit Set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F}

While we are more accustomed to decimal notation for writing numeric values, we find that
they can be written in any base.  Here are the binary, octal, and hexadecimal equivalents of
the first seventeen non–negative decimal integers

      Decimal      Binary     Octal      Hexadecimal
                        (base 2)   (base 8)        (base 16)

           0             0000         00                  0

           1             0001         01                  1                 Note that conversions from hexadecimal

           2             0010         02                  2                 to binary can be done one digit at a time,

           3             0011         03                  3                 thus DE = 11011110, as D = 1101 and

           4             0100         04                  4                 E = 1110.  We shall normally denote

           5             0101         05                  5                 this as DE = 1101 1110 with a space

           6             0110         06                  6                 to facilitate reading the binary.

           7             0111         07                  7

           8             1000         10                  8                 Conversion from binary to hexadecimal

           9             1001         11                  9                 is also quite easy.  Group the bits four at

          10            1010         12                  A                 a time and convert each set of four.

          11            1011         13                  B                 Thus 10111101, written 1011 1101 for

          12            1100         14                  C                 clarity is BD because 1011 = B and

          13            1101         15                  D                 1101 = D.

          14            1110         16                  E

          15            1111         17                  F
          16           10000        20                 10


 

Notation for Constants

Given the choice of at least four possible values for the base of the numbering system
(2, 8, 10, and 16), how do we indicate which base is to be used to interpret a constant
value, such as 011?  This is a valid number in each of the bases, though its value
depends on which base is being used.

Pure binary numbers will rarely be used in these notes.  When they are, they will be
almost obvious, having only 0 or 1 as digits.  These constants will be noted by a small
subscript 2, as in 0112, the binary equivalent of decimal 3.

For values written in octal, decimal, and hexadecimal notation, these notes, along with
most other textbooks, use the notation of Java and C/C++ to indicate the choice of base
values.  Decimal values are written simply as one would expect.  Octal constants are
prefixed by a leading “0” (zero), while hexadecimal constants are prefixed by “0x”.

For example:
      value = 77     // This is the decimal value 77.  No special notation is used.
      value = 077   // This is the octal value 077, equal to 7·8 + 7 = 62 decimal.
      value = 0x77            // This is hexadecimal 77, equal to 7·16 + 7 = 119 decimal.

Positional Notation and Decimal Numbers

Before discussing number bases other than decimal, it would be helpful to use the decimal
system to review positional notation, as always used to write integer values.

Positional notation is formed using powers of the base.  These powers are denoted by
exponents, written as superscripts.  For integer representations, all exponents are taken
from the set of non–negative integers.  We review this notation.

Any non–zero number raised to the zero power has value 1: N0 = 1, for N ≠ 0.
Here are some powers of the various bases. 
      100 = 1                                     20 = 1                           160 = 1
      101 = 10                                   21 = 2                           161 = 24 = 16
      102 = 10·10 = 100                   22 = 2·2 = 4                 162 = 28 = 256
      103 = 10·100 = 1000               23 = 2·4 = 8                 163 = 212 = 4,096
      104 = 10·103 = 10,000            24 = 2·23 = 16             164 = 216 = 65,536

Consider the value1453, commonly understood as “one thousand four hundred fifty three”. 
Specifically 1453 = 1000 + 400 + 50 + 3.  Written in terms of powers of ten, we see that
1453 = 1·103 + 4·102 + 5·101 + 3·100.  Similarly, the larger number 322,583,006 which
represents 3·108 + 2·107 + 2·106 + 5·105 + 8·104 + 3·103 + 0·102 + 0·101 + 6·100.

Positional Notation and Other Base Values

We are now in a position to generalize the idea of positional notation.  Consider 1453.
We have already seen the decimal evaluation of 1453. 
If read as octal notation 01453 = 1·83 + 4·82 + 5·81 + 3·80 = 811 decimal.
If read as hexadecimal notation 0x1453 = 1·163 + 4·162 + 5·161 + 3·160 = 5,203 decimal.
The value 1453 cannot be read as binary, because it has digits other than 0 and 1.


 

Which Base to Use?

The choice of the base to use is made solely for the convenience of the person who must
examine or otherwise use the information.  In many courses, this person is the instructor
who must grade the homework and who prefers a more compact notation.

Almost always, decimal notation is preferable when writing a computer program in any
language, such as COBOL, Java, C/C++, or even IBM assembler.  The examples below,
which illustrate proper usage of other bases, are given without explanation.

1.   Sometimes it is preferable to use hexadecimal notation when writing software, known as
      a device driver, used to control external devices, such as printers and disks.
      For example, one might write: device_mask = 0xC040.

      This line of code will set binary bits in a device control register.  The register is seen to
      have sixteen bits, with values set as follows:

Bit

15

14

13

12

11

10

9

8

7

6

5

4

3

2

1

0

Value

1

1

0

0

0

0

0

0

0

1

0

0

0

0

0

0

      Of course, one could just as easily write this as device_mask = –16320 using decimal
      notation, but most programmers would prefer the hexadecimal.  One might note here that,
      with the exception of classroom exercises, very few programmers write device drivers.

2.   Some time it is necessary to examine the raw contents of memory.  Thankfully, this is rarely
necessary as high–grade debuggers do the job for us.  Again, the most common use
      for this these days is the assignment of vexing problems by overly zealous professors.

      As an example (to be explained later), here is the binary representation of the real number
      –0.75: 1011 1111 0100 0000 0000 0000 0000 0000

      Note that the 32 binary bits are grouped by fours for ease of reading.  The hexadecimal
      notation is even easier to read: BF40 0000.

3.   The choice of octal or hexadecimal notation usually depends on the value to be
      represented.  Octal digits can be considered as short hand for groups of three binary
      bits, while hexadecimal digits are short hand for groups of four binary bits.

      Most values to be represented in either binary or a variant of it are best considered as
      values having 8, 16, 32, or 64 bits.  As each of these lengths is a multiple of four, the
      most natural choice would be hexadecimal notation.  Consider the example above.

      Some older, and now obsolete, computers had word lengths that were multiples of 3;
      for these octal notation might be preferred.  The old PDP–8 used 12–bit words, and
      the old PDP–9 used 18–bit words, represented as 4 or 6 octal digits respectively.

Note on History:     This author is quite fond of mentioning obsolete computers, among
                                 them the PDP–8, PDP–9, PDP–11, VAX–11/780, CDC–6600,
                                 CDC–7600, the Cray–1, and Cray–2.  This is done to show that there
                                 were actual computers with such–and–such properties.
                                 Nothing in this book requires any knowledge of such old favorites.

Note on Usage:       The remainder of this book will usually ignore the octal base.


 

The fact that the bases for octal and hexadecimal are powers of the basis for binary
facilitates the conversion between the bases 2, 8, and 16.  The conversion can be done
one digit at a time, remembering that each octal digit corresponds to three binary bits
and each hexadecimal digit corresponds to four binary bits.  Conversion between decimal
and one of the other three bases is only a bit more complex.

As noted above, we shall focus on binary, decimal, and hexadecimal numbers.  As octal
notation is rarely used these days, we shall not consider it further.

How to Group the Bits?

This is a small problem that occurs in translations between binary and hexadecimal notation. 
As each hexadecimal digit stands for four binary bits, these two translations (binary to hex
and hex to binary) are just matters of grouping bits.  We begin by discussing integers.

Consider the hexadecimal number BAD.  We translate digit by digit to obtain the binary. 
Hexadecimal B stands for binary 1011, A for 1010, and D for 1101.  The binary value
represented is 101110101101, better written as 1011 1010 1101.

Consider conversion of the binary number 111010 to hexadecimal.  If we try to group the
bits four at a time we get either 11 1010 or 1110 10.  The first option is correct as the
grouping must be done from the right.  We then add leading zeroes to get groups of four
binary bits, obtaining 0011 1010, which is converted to 3A as 0011 = 3 and 1010 = A.
Note that the padding of the leading 11 to the full four–bit 0011 is not strictly necessary,
as most readers are quite comfortable with interpreting binary 11 as decimal 3.

We shall have occasion to do conversions involving binary and hexadecimal to the right
of the decimal point.  Consider the hexadecimal value 0.C8, which is 0.1100 1000 in
binary.  Here the grouping is not a problem; just represent each hex digit as four bits.

Consider the binary value 0.111010, which represents the decimal value 0.90625
(think that 0.111010 stands for 1/2 + 1/4 + 1/8 + 0/16 + 1/32 = 29/32).  When the digits
are to the right of the decimal, they must be grouped from the left and padded out with
trailing zeroes so that each collection has four bits.  Thus 0.111010 is converted to
 0.1110 1000, or 0.C8.  Note that leaving the value as 0.1110 10 might lead
one to consider the last grouping as decimal 2, which is likely to cause confusion.

Unsigned Binary Integers

There are two common methods to store unsigned integers in a computer: binary numbers
(which we discuss now) and Packed Decimal (which we discuss later).  From a theoretical
point of view, it is important to note that no computer really stores the set of integers in
that it can represent in native form an arbitrary member of that infinite set. 

For each of the standard formats (short, integer, long, etc.) there is a largest positive value
that can be stored and a smallest non–positive value that can be stored.  More on this later.
There is a common data type, usually called “bignum”, which allows storage of integers of
any size.  You want a 10,000–digit integer, you got one.  However, this data type is mostly
implemented in software with hardware assists, and is not a native type of a standard CPU.


 

The standard integer types are stored as collections of a number of binary bits; thus an
N–bit integer is represented by N binary bits.  1110 1000 would be an 8–bit integer.

It is easy to show that an N–bit binary integer can represent one of 2N possible integer values. 
Here is the proof by induction.

      1.   A one–bit integer can store 2 values: 0 or 1.  This is the base for induction.

      2.   Suppose an N–bit integer, unconventionally written as BN–1 … B3B2B1B0.
            By the inductive hypothesis, this can represent one of 2N possible values.

      3.   We now consider an (N+1)–bit integer, written as BNBN–1 … B3B2B1B0,
            where either BN = 0 or BN = 1.  By the inductive hypothesis, there are
                     2N values of the form 0BN–1 … B3B2B1B0,
            and   2N values of the form 1BN–1 … B3B2B1B0.

      4.   The total number of (N+1)–bit values is 2N + 2N = 2N+1.  The claim is proved.

By inspection of the above table, we see that there are 16 possible values for a four–bit
unsigned integer.  These range from decimal 0 through decimal 15 and are easily represented
by a single hexadecimal digit.  Each hexadecimal digit is shorthand for four binary bits.

In the standard interpretation, always used in this course, an N–bit unsigned integer will
represent 2N integer values in the range 0 through 2N – 1, inclusive.  Sample ranges include:

      N =          4      0 through     24 – 1         0 through                          15
      N =          8      0 through     28 – 1         0 through                        255
      N =        12      0 through    212 – 1         0 through                      4095
      N =        16      0 through    216 – 1         0 through                    65535
      N =        20      0 through    220 – 1         0 through              1,048,575
      N =        32      0 through    232 – 1         0 through       4,294,967,295

For most applications, the most important representations are 8 bit, 16 bit, and 32 bit.  To this
mix, we add 12–bit unsigned integers as they are used in the base register and offset scheme
of addressing used by the IBM Mainframe computers.  Recalling that a hexadecimal digit is
best seen as a convenient way to write four binary bits, we have the following.

      8 bit numbers        2 hexadecimal digits      0 through                          255,
      12 bit numbers      3 hexadecimal digits      0 through                        4095,
      16 bit numbers      4 hexadecimal digits      0 through                      65535,         and
      32 bit numbers      8 hexadecimal digits      0 through         4,294,967,295.

Conversions between Decimal and Binary

We now consider methods for conversion from decimal to binary and binary to decimal. We
consider not only whole numbers (integers), but numbers with decimal fractions.  To convert
such a number, one must convert the integer and fractional parts separately.

At this point in the text, we have yet to consider signed integers.  The reason for the focus on
unsigned values is that all conversions between decimal and the other two formats (binary
and hexadecimal) are best learned using unsigned values.


 

Consider the conversion of the number 23.375.  The method used to convert the integer part
(23) is different from the method used to convert the fractional part (.375).  We shall discuss
two distinct methods for conversion of each part and leave the student to choose his/her
favorite.  After this discussion we note some puzzling facts about exact representation of
decimal fractions in binary; e.g. the fact that 0.20 in decimal cannot be exactly represented. 
As before we present two proofs and let the student choose his/her favorite.

The intuitive way to convert decimal 23 to binary is to note that 23 = 16 + 7 = 16 + 4 + 2 + 1;
thus decimal 23 = 10111 binary.  As an eight bit binary number, this is 0001 0111.  Note that
we needed 5 bits to represent the number; this reflects the fact that 24 < 23 £ 25.  We expand
this to an 8-bit representation by adding three leading zeroes.

The intuitive way to convert decimal 0.375 to binary is to note that 0.375 = 1/4 + 1/8 =
0/2 + 1/4 + 1/8, so decimal .375 = binary .011 and decimal 23.375 = binary 10111.011.

Most students prefer a more mechanical way to do the conversions.  Here we present that
method and encourage the students to learn this method in preference to the previous.

Conversion of integers from decimal to binary is done by repeated integer division with
keeping of the integer quotient and noting the integer remainder.  The remainder numbers are
then read bottom to top as least significant bit to first bit generated.  Here is an example.

                           Quotient    Remainder

             23/2 =        11                  1                 Thus decimal 23 = binary 10111

             11/2 =         5                   1

              5/2 =          2                   1                 Remember to read the binary number

              2/2 =          1                   0                 from bottom to top.

              1/2 =          0                   1

Conversion of the fractional part is done by repeated multiplication with copying of the
whole number part of the product and subsequent multiplication of the fractional part.  All
multiplications are by 2.  Read the binary results top to bottom.  Here is an example.

                        Number           Product     Binary

                         0.375     x 2 =     0.75          0

                         0.75       x 2 =     1.5            1

                         0.5         x 2 =     1.0            1
                         0.0         x 2 =     0.0            0
                         0.0         x 2 =     0.0            0

The process terminates when the product of the last multiplication is 1.0.  At this point we
copy the last 1 generated and have the result; thus decimal 0.375 = 0.011 binary.  Note that
this conversion was carried two steps beyond the standard termination point to emphasize
the fact that after a point, only trailing zeroes are generated.

0.375 = 0.3750 = 0.37500, etc.  0.011 = 0.0110 = 0.01100, etc.


 

Conversions between Decimal and Hexadecimal

One way to convert is by first converting to binary.  We consider conversion of 23.375
from decimal to hexadecimal.  We have noted that the value is 10111.011 in binary.

 

To convert this binary number to hexadecimal we must group the binary bits in groups of
four, adding leading and trailing zeroes as necessary.  We introduce spaces in the numbers
in order to show what is being done.

         10111.011 = 1 0111.011.

To the left of the decimal we group from the right and to the right of the decimal we group
from the left.  Thus 1.011101 would be grouped as 1.0111 01.

At this point we must add extra zeroes to form four bit groups.  So

         10111.011 = 0001 0111.0110.

Conversion to hexadecimal is done four bits at a time.  The answer is 17.6 hexadecimal.

Another Way to Convert Decimal to Hexadecimal

Some readers may ask why we avoid the repeated division and multiplication methods in
conversion from decimal to hexadecimal.  Just to show it can be done, here is an example.

Consider the number 7085.791748046875.  As an example, we convert this to hexadecimal.

The first step is to use repeated division to produce the whole–number part.

         7085 / 16         = 442          with remainder = 13            or hexadecimal D

           442 / 16           = 27          with remainder = 10            or hexadecimal A
             27 / 16             = 1          with remainder = 11            or hexadecimal B
               1 / 16             = 0            with remainder = 1            or hexadecimal 1.

The whole number is read bottom to top as 1BAD.

Now we use repeated multiplication to obtain the fractional part.

    0.791748046875 · 16 =        12.6679875      Remove the 12      or hexadecimal C

              0.6679875 · 16 =        10.6875            Remove the 10      or hexadecimal A

                    0.6875 · 16 =        11.00                Remove the 11      or hexadecimal B

                        0.00 · 16 =          0.0

The fractional part is read top to bottom as CAB.  The hexadecimal value is 1BAD.CAB,
which is a small joke on the author’s part.  The only problem is to remember to write
results in the decimal range 10 through 15 as hexadecimal A through F.

Long division is of very little use in converting the whole number part.  It does correctly
produce the first quotient and remainder.  The intermediate numbers may be confusing.

       442

   16)7085

      64

       68

       64

        45

        32

        13


 

But Why Do These Methods Work?

In this optional section, we present some illustrative examples to indicate why each of these
conversion methods work.

Conversion of Integers: Decimal to Binary

Our sample integer, used above, is 23 = 16 + 4 + 2 + 1 = 1·24 + 0·23 + 1·22 + 1·2 + 1.  To
generalize this to any number representable as 5–bit unsigned binary, we use some algebra.
Let 23 = A4·24 + A3·23 + A2·22 + A1·2 + A0, where A4 = 1, A3 = 0, A2 = 1, etc.  Now begin
the repeated divisions by 2, noting that we track only whole numbers.

(A4·24 + A3·23 + A2·22 + A1·2 + A0) / 2   = (A4·23 + A3·22 + A2·2 + A1),        rem = A0.
(A4·23 + A3·22 + A2·2 + A1) / 2                 = (A4·22 + A3·2 + A2),                      rem = A1.
(A4·22 + A3·2 + A2) / 2                              = (A4·2 + A3),                                   rem = A2.
(A4·2 + A3) / 2                                            = A4                                                   rem = A3.
A4 / 2                                                          = 0                                                     rem = A4.

Continuation of the process will generate only leading zeroes, reflecting the fact that the
decimal value 23 is binary 10111 = 010111 = 0010111 = 0001 0111, etc.

Conversion of “Fractional Parts”: Decimal to Binary

Our sample “fractional part” (the author does not know a better term) is 0.375.  Since this
conversion moves a bit quickly, we use here another value: 0.65625 = 1/2 + 1/8 + 1/32. 
Let 0.65625 = B1/2 + B2/22 + B3/23 + B4/24 + B5/25, with B1 = 1, B2 = 0, B3 = 1, etc.
Now begin the repeated multiplications, tracking only the “fractional part”.

(B1/2 + B2/22 + B3/23 + B4/24 + B5/25) · 2       = B1 + (B2/2 + B3/22 + B4/23 + B5/24)
(B2/2 + B3/22 + B4/23 + B5/24) · 2                   = B2 + (B3/2 + B4/22 + B5/23)
(B3/2 + B4/22 + B5/23) · 2                                = B3 + (B4/2 + B5/22)
(B4/2 + B5/22) · 2                                            = B4 + B5/2
B5/2 · 2                                                           = B5

Continuing this will lead to trailing zeroes, the value in binary being either 0.10101 or
0.101010, or more conventionally 0.1010 1000.

What about Other Bases?

The methods above apply to conversion between any two radix values, though they are
probably difficult to use for conversions from any base other than decimal (as we are quite
familiar with decimal arithmetic).  For us, the only other base is hexadecimal.

Consider our sample number 7085 = A3·163 + A2·162 + A1·16 + A0, with A3 = 1,
A2 = 11 (B in hexadecimal), A1 = 10 (0xA), and A0 = 13 (0xD).

(A3·163 + A2·162 + A1·16 + A0) / 16       = (A3·162 + A2·16 + A1),             rem = A0.
(A3·162 + A2·16 + A1) / 16                       = (A3·16 + A2),                             rem = A1.
(A3·16 + A2) / 16                                      = A3                                                              rem = A2.
A3 / 16                                                       = 0,                                                rem = A3.

 


Non-terminating Fractions

We now make a detour to note a surprising fact about binary numbers – that some fractions
that terminate in decimal are non-terminating in binary.  We first consider terminating and
non-terminating fractions in decimal.  All of us know that 1/4 = 0.25, which is a terminating
fraction, but that 1/3 = 0.333333333333333333333333333333…, a non-terminating fraction.

We offer a demonstration of why 1/4 terminates in decimal notation and 1/3 does not, and
then we show two proofs that 1/3 cannot be a terminating fraction in either binary or decimal.

Consider the following sequence of multiplications
         ¼ · 10 = 2½
         ½ · 10 = 5.  Thus 1/4 = 25/100 = 0.25.

Put another way, ¼ = (1/10) · (2 + ½) =  (1/10) · (2 + (1/10) · 5).

However, 1/3 · 10 = 10/3 = 3 + 1/3, so repeated multiplication by 10 continues to yield a
fraction of 1/3 in the product; hence, the decimal representation of 1/3 is non-terminating.

Explicitly, we see that 1/3 = (1/10) · (3 + 1/3) = (1/10) · (3 + (1/10) · (3 + 1/3)), etc.

In decimal numbering, a fraction is terminating if and only if it can be represented in the
form J / 10K for some integers J and K.  We have seen that 1/4 = 25/100 = 25/102, thus the
fraction 1/4 is a terminating fraction because we have shown the integers J = 25 and K = 2.

Here are two proofs that the fraction 1/3 cannot be represented as a terminating fraction in
decimal notation.  The first proof relies on the fact that every positive power of 10 can be
written as 9·M + 1 for some integer M.  The second relies on the fact that 10 = 2·5, so that
10K = 2K·5K.  To motivate the first proof, note that 100 = 1 = 9·0 + 1, 10 = 9·1 + 1,

100 = 9·11 + 1, 1000 = 9·111 + 1, etc.  If 1/3 were a terminating decimal, we could solve
the following equations for integers J and M.

, which becomes 3·J = 9·M + 1 or 3·(J – 3·M) = 1.  But there is no
integer X such that 3·X = 1 and the equation has no integer solutions.

The other proof also involves solving an equation.  If 1/3 were a non-terminating fraction,
then we could solve the following equation for J and K.

, which becomes 3·J = 2K·5K.  This has an integer solution J only if the
right hand side of the equation can be factored by 3.  But neither 2K nor 5K can be factored by
3, so the right hand side cannot be factored by 3 and hence the equation is not solvable.

From this, it immediately follows that the fraction 1/3, as a decimal number, does not
terminate.  The fact that the base–3 representation (0.13), base–6 representation (0.23), and
base–9 representation (0.39) do terminate is only of marginal academic interest.

NOTE:     This discussion of an obscure point in numeric representation might seem to be
                  pointless, but it has direct bearing on the precision with which real numbers may
                  be stored and processed in a computer.


Now consider the innocent looking decimal 0.20.  We show that this does not have a
terminating form in binary.  We first demonstrate this by trying to apply the
multiplication method to obtain the binary representation.

 

         Number           Product     Binary

         0.20 · 2 =           0.40           0
         0.40 · 2 =           0.80           0

         0.80 · 2 =           1.60           1

         0.60 · 2 =           1.20           1

         0.20 · 2 =           0.40           0

         0.40 · 2 =           0.80           0

         0.80 · 2 =           1.60           1    but we have seen this – see four lines above.

So decimal 0.20 in binary is 0.00110011001100110011 …, ad infinitum.  This might
be written conventionally as 0.00110 0110 0110 0110 0110, to emphasize the pattern.

The proof that no terminating representation exists depends on the fact that any terminating
fraction in binary can be represented in the form  for some integers J and K.  Thus we
solve  or 5·J = 2K.  This equation has a solution only if the right hand side is divisible
by 5.  But 2 and 5 are relatively prime numbers, so 5 does not divide any power of 2 and the
equation has no integer solution.  Hence 0.20 in decimal has no terminating form in binary.

 

Binary Addition

The next topic is storage of integers in a computer.  We shall be concerned with storage of
both positive and negative integers.  Two’s complement arithmetic is the most common
method of storing signed integers.  Calculation of the two’s complement of a number
involves binary addition.  For that reason, we first discuss binary addition.

To motivate our discussion of binary addition, let us first look at decimal addition.  Consider
the sum 15 + 17 = 32.  First, note that 5 + 7 = 12.  In order to speak of binary addition, we
must revert to a more basic way to describe 5 + 7; we say that the sum is 2 with a carry-out of
1.  Consider the sum 1 + 1, which is known to be 2.  However, the correct answer to our
simple problem is 32, not 22, because in computing the sum 1 + 1 we must consider the
carry-in digit, here a 1.  With that in mind, we show two addition tables – for a half-adder
and a full-adder.  The half-adder table is simpler as it does not involve a carry-in.  The
following table considers the sum and carry from A + B.

Half-Adder A + B

        A   B       Sum    Carry          

        0    0          0          0              Note the last row where we claim that 1 + 1 yields a

        0    1          1          0              sum of zero and a carry of 1.  This is similar to the

        1    0          1          0              statement in decimal arithmetic that 5 + 5 yields a

        1    1          0          1              sum of 0 and carry of 1 when 5 + 5 = 10.

 

Remember that when the sum of two numbers equals or exceeds the value of the base of the
numbering system (here 2) that we decrease the sum by the value of the base and generate a
carry.  Here the base of the number system is 2 (decimal), which is 1 + 1, and the sum is 0.
Say “One plus one equals two plus zero: 1 + 1 = 10”.

 

For us the half-adder is only a step in the understanding of a full-adder, which implements
binary addition when a carry-in is allowed.  We now view the table for the sum A + B, with
a carry-in denoted by C.  One can consider this A + B + C, if that helps.

 

Full-Adder: A + B with Carry

                 A      B      C                Sum       Carry

                 0       0       0                   0             0

                 0       0       1                   1             0

                 0       1       0                   1             0

                 0       1       1                   0             1

                 1       0       0                   1             0

                 1       0       1                   0             1

                 1       1       0                   0             1

                 1       1       1                   1             1

In the next chapter, we shall investigate the construction of a full adder from digital
gates used to implement Boolean logic.  Just to anticipate the answer, we note that the
sum and carry table above is in the form of a Boolean truth table, which can be
immediately converted to a Boolean expression that can be implemented in digital logic.

As an example, we shall consider a number of examples of addition of four-bit binary
numbers.  The problem will first be stated in decimal, then converted to binary, and then
done.  The last problem is introduced for the express purpose of pointing out an error.

 

We shall see in a minute that four-bit binary numbers can represent decimal numbers in
the range 0 to 15 inclusive.  Here are the problems, first in decimal and then in binary.

      1)      6 + 1             0110 + 0001

      2)      11 + 1           1011 + 0001

      3)      13 + 5           1101 + 0101

 

  0110     1011     1101      In the first sum, we add 1 to an even number.  This

  0001     0001     0101        is quite easy to do.  Just change the last 0 to a 1.

  0111     1100     0010      Otherwise, we may need to watch the carry bits.

 

In the second sum, let us proceed from right to left.  1 + 1 = 0 with carry = 1.  The second
column has 1 + 0 with carry-in of 1 = 0 with carry-out = 1.  The third column has 0 + 0
with a carry-in of 1 = 1 with carry-out = 0.  The fourth column is 1 + 0 = 1.

Analysis of the third sum shows that it is correct bit-wise but seems to be indicating that
13 + 5 = 2.  This is an example of “busted arithmetic”, more properly called overflow. 
A given number of bits can represent integers only in a given range; here 13 + 5 is outside
the range 0 to 15 inclusive that is proper for four-bit unsigned numbers.

Signed and Unsigned Integers

Up to this point, we have discussed only unsigned integers and the conversion of such from
one base to another.  We now consider signed integers and find that consideration of the
magnitude of such numbers as unsigned integers is an important part of the process.

Integers are stored in a number of formats.  The most common formats today include 16 and
32 bits, though long integers (64–bits) are becoming a standard integer format. 
Although 32-bit integers are probably the most common, our examples focus on eight–bit
integers because they are easy to illustrate.  In these discussions, the student should recall the
powers of 2: 20 = 1, 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64, 27 = 128, and 28 = 256.

Bits in the storage of an integer are numbered right to left, with bit 0 being the right-most or
least-significant.  In eight bit integers, the bits from left to right are numbered 7 to 0.  In 32
bit integers, the bits from left to right are numbered 31 to 0.  Note that this is not the notation
used by IBM for its mainframe and enterprise computers. In the IBM notation, the most
significant bit (often the sign bit) is bit 0 and the least significant bit has the highest number;
bit 7 for an 8–bit integer.  Here are the bit numberings for a signed 8–bit integer.

Common Notation (8–bit entry)                                    IBM Mainframe Notation

Bit #

7

6

5 – 1

0

 

Bit #

0

1

2 – 6

7

 

Sign

MSB

 

LSB

 

Sign

MSB

 

LSB

The simplest topic to cover is the storage of unsigned integers.  As there are 2N possible
combinations of N binary bits, there are 2N unsigned integers ranging from 0 to 2N – 1.  For
eight-bit unsigned integers, this range is 0 though 255, as 28 = 256.  Conversion from binary
to decimal is easy and follows the discussion earlier in this chapter.

Of the various methods for storing signed integers, we shall discuss only three
      Two’s complement
      One’s complement (but only as a way to compute the two’s complement)
      Excess 127 (for 8–bit numbers only) as a way to understand the floating point standard.

One’s complement arithmetic is mostly obsolete and interests us only as a stepping-stone to
two’s complement arithmetic.  To compute the one’s complement of a number:

            1) Represent the number as an N-bit binary number
            2) Convert every 0 to a 1 and every 1 to a 0.

Decimal 100 =             0110 0100

One’s complement      1001 1011; decimal –100 = 1001 1011 binary.

But consider the value 0 and problems representing it in one’s–complement notation.
Decimal 0 =                 0000 0000

One’s complement      1111 1111; decimal –0 = 1111 1111 binary.

There are a number of problems in one’s complement arithmetic, the most noticeable being
illustrated by the fact that the one’s complement of 0 is 1111 1111.  In this system, we have
–0 ¹ 0, which is a violation of some of the basic principles of mathematics.  When coding
for a computer that used one’s–complement arithmetic, one had to write code such as:
            if ( (x == 0) || (x == –0) ) …


 

Notation:

In discussing the one’s complement notation, we shall borrow some notation from digital
logic.  This notation will be fully discussed in a later chapter, but is quite easy to grasp.

We introduce the idea of logical NOT, as applied to binary values 0 and 1.  Simply put
            NOT(0) = 1
            NOT(1) = 0.

The logical NOT function is denoted in three distinct ways, but here we focus on only one:
the over–bar notation.  Thus = 1 and = 0.

Let X represent an integer value.  Note that the recipe for taking the one’s–complement of a
number involves taking the logical NOT of each of its binary bits.  For this reason, I propose
to denote the one’s–complement of an integer X by , using the NOT symbol.

Let X = 100.   Represented in binary,            X =      0110 0100
         
The one’s complement is         =   1001 1011

Note that for addition X + = 1111 1111.  We shall say more on this later.

The Two’s Complement
For integer arithmetic, all modern computers use two’s–complement notation.  In this text,
we shall reserve the standard symbol “–X” to denote the negative in the two’s–complement
system.  The two’s complement of a number is obtained as follows:

            1) First take the one’s complement of the number

            2) Add 1 to the one’s complement and discard the carry out of the left-most column.

Decimal 100 =             0110 0100

One’s complement      1001 1011

We now do the addition         1001 1011
                                                        1
                                                1001 1100

Thus, in eight-bit two’s complement arithmetic

      Decimal 100          = 0110 0100 binary
      Decimal – 100       = 1001 1100 binary

This illustrates one pleasing feature of two’s complement arithmetic: for both positive and
negative integers the last bit is zero if and only if the number is even.  Note that it is essential
to state how many bits are to be used.  Consider the 8-bit two’s complement of 100.  Now
100 = 64 + 32 + 4, so decimal 100 = 0110 0100 binary, and we get the result above.

Consider decimal 12 = 0000 1100 binary.  If we took the two’s complement of 1100, we
might get 0100, giving us no idea how to pad out the high–order four bits.

The real reason for the popularity of two’s complement can be seen by calculating the
representation of – 0.  To do this we take the two’s complement of 0.


 

In eight bits, 0 is represented as                0000 0000

Its one’s complement is represented as     1111 1111.

We now take the two’s complement of 0.

Here is the addition      1111 1111
                                              1
                                    1 0000 0000 – but discard the leading 1.

Thus the two’s complement of 0 is represented as 0000 0000.  As required by algebra,
this is exactly the same as the representation of 0 and we avoid the messy problem – 0 ¹ 0.

In 8– bit two’s complement arithmetic,                + 127         0111 1111
the range of integers that can be                           + 10           0000 1010
represented is – 2N-1 through 2N-1 – 1                    +1              0000 0001
inclusive, thus the range for eight-bit                   0                0000 0000
two’s complement integers is –128                       – 1             1111 1111  The number is
through 127 inclusive, as 27 = 128.  The               – 10           1111 0110  negative if and
table at the right shows a number of                     – 127         1000 0001  only if the left–
binary representations for this example.               – 128         1000 0000  most bit is 1.

 

We now give the ranges allowed for the most common two’s complement representations.
      Eight bit                – 128         to                      +127
      16-bit                – 32,768         to                 +32,767
      32-bit    – 2,147,483,648         to     +2,147,483,647

 

The range for 64-bit two’s complement integers is – 263 to 263 – 1.  As an exercise in math,
I propose to do a rough calculation of 263.  This will be done using only logarithms.

There is a small collection of numbers that the serious student of computer science should
memorize.  Two of these numbers are the base-10 logarithms of 2 and 3.  To five decimal
places, log 2 = 0.30103 and log 3 = 0.47712.

Now 263 = (100.30103)63 = 1018.9649 = 100.9649 · 1018.  What do we make of 100.9649?  Since we
have log 3 = 0.47712 and 9 = 32, we have log 9 = 2 · log 3 = 2 · 0.47712 = 0.95424.  This
leads to the conclusion that 100.9649 > 100.95424 » 9; hence 9. < 100.95424 < 10.0.  We conclude
that a 64–bit integer allows the representation of 18 digit numbers and most 19 digit values.
The precise range is –9,223,372,036,854,775,808 through 9,223,372,036,854,775,807.

 

Reminder: For any number of bits, in two’s complement arithmetic the number is negative if
and only if the high-order bit in the binary representation is a 1.

 

Sign Extension

This applies to numbers represented in one’s-complement and two’s-complement form.  The
issue arises when we store a number in a form with more bits; for example when we store a
16-bit integer in a 32-bit register.  The question is how to set the high-order bits.

 

Consider a 16-bit integer stored in two’s-complement form.  Bit 15 is the sign bit.  We can
consider bit representation of the number as A15A14A13A12A11A10A9A8A7A6A5A4A3A2A1A0. 
Consider placing this number into a 32-bit register with bits numbered R31 through R0, with
R31 being the sign bit.  Part of the solution is obvious: make Rk = Ak for 0 £ k £ 15.  What is
not obvious is how to set bits 31 through 16 in R as the 16-bit integer A has no such bits.

 

For non-negative numbers the solution is obvious and simple, set the extra bits to 0.  This is
like writing the number two hundred (200) as a five digit integer; write it 00200.  But
consider the 16-bit binary number 1111 1111 1000 0101, decimal –123. 
If we expanded this to 0000 0000 0000 0000 1111 1111 1000 0101 by setting
the high order bits to 0’s, we would have a positive number, 65413.  This is not correct.

 

The answer to the problem is sign extension, which means filling the higher order bits of the
bigger representation with the sign bit from the more restricted representation.  In our
example, we set bits 31 through 16 of the register to the sign bit of the 16-bit integer. 
The correct answer is then 1111 1111 1111 1111 1111 1111 1000 0101.

 

Note – the way I got the value 1111 1111 1000 0101 for the 16-bit representation of – 123
was to compute the 8-bit representation, which is 1000 0101.  The sign bit in this
representation is 1, so I extended the number to 16-bits by setting the high order bits to 1.

Nomenclature: “Two’s-Complement Representation” vs. “Taking the Two’s-Complement”

We now address an issue that seems to cause confusion to some students.  There is a
difference between the idea of a complement system and the process of taking the
complement.  Because we are interested only in the two’s-complement system, I restrict my
discussion to that system.

 

Question: What is the representation of the positive number 123 in 8-bit two’s complement
arithmetic?

Answer: 0111 1011.  Note that I did not take the two’s complement of anything to get this.

 

Two’s-complement arithmetic is a system of representing integers in which the two’s-
complement is used to compute the negative of an integer.  For positive integers, the method
of conversion to binary differs from unsigned integers only in the representable range.

 

For N-bit unsigned integers, the range of integers representable is 0 ... 2N – 1, inclusive.  For
N-bit two’s-complement integers the range of non-negative integers representable is

0 ... 2N-1 – 1, inclusive.  The rules for converting decimal to binary integers are the same for
non-negative integers – one only has to watch the range restrictions.

 

The only time when one must use the fact that the number system is two’s-complement
(that is – take the two’s-complement) is when one is asked about a negative number.  Strictly
speaking, it is not necessary to take the two’s-complement of anything in order to represent a
negative number in binary, it is only that most students find this the easiest way.

 


 

Question: What is the representation of –123 in 8-bit two’s-complement arithmetic?

Answer: Perhaps I know that the answer is 1000 0101.  As a matter of fact, I can calculate
this result directly without taking the two’s-complement of anything, but most students find
the mechanical way the easiest way to the solution.  Thus, the preferred solution for most students is

            1)         We note that 0 £ 123 £ 27 – 1, so both the number and its negative
                        can be represented as an 8-bit two’s-complement integer.

            2)         We note that the representation of +123 in 8-bit binary is 0111 1011

            3)         We take the two’s-complement of this binary result to get the binary
                        representation of –123 as 1000 0101.

We note in passing a decidedly weird way to calculate the representations of non-negative
integers in two’s-complement form.  Suppose we want the two’s-complement representation
of +123 as an eight-bit binary number.  We could start with the fact that the representation
of –123 in 8-bit two’s-complement is 1000 0101 and take the two’s complement of 1000 0101
to obtain the binary representation of 123 = –(–123).  This is perfectly valid, but decidedly strange. 
One could work this way, but why bother?

Summary:      Speaking of the two’s-complement does not mean that one must take the
                        two’s-complement of anything.

Why Does the Two’s Complement Work?

We now ask an obvious question.  The process of taking the two’s–complement of the binary
representation of an integer has been described and is well defined.  But why does this
process yield the negative of the integer.  We begin with a fact noted in passing just above

Question:        Name the only number with the property that if I add 1 to it, I get 0.

Answer:          That unique number is negative 1, denoted as –1.

Look now at simple addition for each of 8–bit and 16–bit signed integer values.

In 8–bit        1111 1111
               +            1
                     0000 0000     Remember that the carry–out from the left bit is discarded.

In 16–bit      1111 1111 1111 1111
               +                      1
                     0000 0000 0000 0000    Again, the sum is 0.

From these two examples, it is possible to generalize to a statement that in any system in
which the number 0 is represented as all bits being 0, a number with all bits 1 is –1.

We now consider the bitwise addition of a binary bit and its one’s–complement, also
called NOT(a) or .  Thus we compute the sum a +  for the two possible values of a. 
At the bit level, the addition table is simple, as the sum is always 1, without a carry.

           a                     a +
           0          1                1
           1          0                1


 

We now consider the sum of an arbitrary N–bit binary number and its one’s complement,
recalling that taking the one’s complement is a bit–by–bit operation.

Let A be the N–bit binary number represented as         an–1an–2 … a2a1a0.
Let be the 1’s–complement of A, represented as    .

Positional Notation
We now review the idea of positional notation for binary representations of integers,
beginning with N–bit unsigned integers.  While the discussions are valid for all N > 0,
we shall choose to illustrate with N = 8.

But we have shown that for every bit index that

Thus, we have A + = –1.  This should be clarified by a few examples.  Note that in each
example we add either 0 + 1 or 1 + 0, so there are no carry bits to consider.

Consider the 8–bit representation of the decimal number 100.
Let A         = 0110 0100
then     = 1001 1011
A
+      = 1111 1111, which represents –1.

Consider the 8–bit representation of the negative decimal number –123.

Let A         = 1000 0101
then     = 0111 1010
A
+      = 1111 1111, which represents –1.

If we have A + = –1 for any binary value A, then we have A + ( + 1) = 0.
Hence ( + 1) = –A, the negative of the value A.  This is why two’s complement works.

Here is a table showing how to count up from the most negative integer.

Decimal

Positive

One’s Complement

Two’s Complement

Comment

8

 

 

1000

–8

7

0111

1000

1001

–8 + 1

6

0110

1001

1010

–8 + 2

5

0101

1010

1011

–8 + 3

4

0100

1011

1100

–8 + 4

3

0011

1100

1101

–8 + 5

2

0010

1101

1110

–8 + 6

1

0001

1110

1111

–8 + 7

The above presents an interesting argument and proof, but it overlooks one essential point. 
How does the hardware handle this?  For any sort of adder, the bits are just that.  There is
nothing special about the high–order bit.  It is just another bit, and not interpreted in any
special way.  To the physical adder, the high–order is just another bit.

We shall discuss this problem when we present the logical design of a binary adder.


Arithmetic Overflow – “Busting the Arithmetic”

We continue our examination of computer arithmetic to consider one more topic – overflow. 
Arithmetic overflow occurs under a number of cases:

            1)         when two positive numbers are added and the result is negative
            2)         when two negative numbers are added and the result is positive
            3)         when a shift operation changes the sign bit of the result.

 

In mathematics, the sum of two negative numbers is always negative and the sum of two
positive numbers is always positive.  The overflow problem is an artifact of the limits on
the range of integers and real numbers as stored in computers.  We shall consider only
overflows arising from integer addition.

 

For two’s-complement arithmetic, the range of storable integers is as follows:

      16-bit      – 215 to 215 – 1    or – 32768             to 32767

      32-bit      – 231 to 231 – 1    or – 2147483648   to 2147483647

 

In two’s-complement arithmetic, the most significant (left-most) bit is the sign bit

 

Overflow in addition occurs when two numbers, each with a sign bit of 0, are added and the
sum has a sign bit of 1 or when two numbers, each with a sign bit of 1, are added and the sum
has a sign bit of 0.  For simplicity, we consider 16-bit addition.  As an example, consider the
sum 24576 + 24576 in both decimal and binary.  Note 24576 = 16384 + 8192 = 214 + 213.

 

              24576               0110 0000 0000 0000
              24576               0110 0000 0000 0000
           – 16384               1100 0000 0000 0000

 

In fact, 24576 + 24576 = 49152 = 32768 + 16384.  The overflow is due to the fact that
49152 is too large to be represented as a 16-bit signed integer.

 

As another example, consider the sum (–32768) + (–32768).  As a 16–bit signed integer,
the sum is 0!

            –32768               1000 0000 0000 0000
            –32768               1000 0000 0000 0000
                      0               0000 0000 0000 0000

 

It is easily shown that addition of a validly positive integer to a valid negative integer cannot
result in an overflow.  For example, consider again 16–bit two’s–complement integer
arithmetic with two integers M and N.  We have 0 £ M £ 32767 and –32768 £ N £ 0.  If
|M| ³ |N|, we have 0 £ (M + N) £ 32767 and the sum is valid.  Otherwise, we have
–32768 £ (M + N) £ 0, which again is valid.

 

Integer overflow can also occur with subtraction.  In this case, the two values (minuend
 and subtrahend) must have opposite signs if overflow is to be possible.


Excess–127

We now cover excess–127 representation.  This is mentioned only because it is required
when discussing the IEEE floating point standard.  In general, we can consider an
excess–M notation for any positive integer M.  For an N–bit excess–M representation, the
rules for conversion from binary to decimal are:

            1) Evaluate as an unsigned binary number

            2) Subtract M.

 

To convert from decimal to binary, the rules are

            1) Add M

            2) Evaluate as an unsigned binary number.

 

In considering excess notation, we focus on eight–bit excess–127 notation.  The range of
values that can be stored is based on the range that can be stored in the plain eight–bit
unsigned standard: 0 through 255.  Remember that in excess–127 notation, to store an integer
N we first form the number N + 127.  The limits on the unsigned eight-bit storage require
that 0 £ (N + 127) £ 255, or – 127 £ N £ 128.

 

As an exercise, we note the eight-bit excess-127 representation of – 5, – 1, 0 and 4.

            – 5 + 127 = 122.          Decimal 122 = 0111 1010 binary, the answer.

             – 1 + 127 = 126.         Decimal 126 = 0111 1110 binary, the answer.

            0 + 127 = 127.             Decimal 127 = 0111 1111 binary, the answer.

            4 + 127 = 131              Decimal 131 = 1000 0011 binary, the answer.

 

We have now completed the discussion of common ways to represent unsigned and signed
integers in a binary computer.  We now start our progress towards understanding the storage
of real numbers in a computer.  There are two ways to store real numbers – fixed point and
floating point.  We focus this discussion on floating point, specifically the IEEE standard for
storing floating point numbers in a computer.

 

Normalized Numbers

The last topic to be discussed prior to defining the IEEE standard for floating point numbers
is that of normalized numbers.  We must also mention the concept of denormalized numbers,
though we shall spend much less time on the latter.

 

A normalized number is one with a representation of the form X · 2P, where 1.0 £ X < 2.0. 
At the moment, we use the term denormalized number to mean a number that cannot be so
represented, although the term has a different precise meaning in the IEEE standard.  First,
we ask a question: “What common number cannot be represented in this form?”

 

The answer is zero.  There is no power of 2 such that 0.0 = X · 2P, where 1.0 £ X < 2.0.  We
shall return to this issue when we discuss the IEEE standard, at which time we shall give a
more precise definition of the denormalized numbers, and note that they include 0.0.  For the
moment, we focus on obtaining the normalized representation of positive real numbers.

 


We start with some simple examples.

         1.0       = 1.0 · 20, thus X = 1.0 and P = 0.

         1.5       = 1.5 · 20, thus X = 1.5 and P = 0.

         2.0       = 1.0 · 21, thus X = 1.0 and P = 1

         0.25     = 1.0 · 2-2, thus X = 1.0 and P = -2

         7.0       = 1.75 · 22, thus X = 1.75 and P = 2

         0.75     = 1.5 · 2-1, thus X = 1.5 and P = -1.

 

To better understand this conversion, we shall do a few more examples using the more
mechanical approach to conversion of decimal numbers to binary.  We start with an example:
9.375 · 10-2 = 0.09375.  We now convert to binary.

 

0.09375  · 2 = 0.1875       0
0.1875    · 2 = 0.375         0                Thus decimal 0.09375 = 0.00011 binary
0.375      · 2 = 0.75           0                or 1.1 · 2-4 in the normalized notation.
0.75        · 2 = 1.5             1
0.5          · 2 = 1.0             1

 

Please note that these representations take the form X · 2P, where X is represented as a
binary number but P is represented as a decimal number.  Later, P will be converted to an
excess–127 binary representation, but for the present it is easier to keep it in decimal.

 

We now convert the decimal number 80.09375 to binary notation.  I have chosen 0.09375 as
the fractional part out of laziness as we have already obtained its binary representation.  We
now convert the number 80 from decimal to binary.  Note 80 = 64 + 16 = 26 · (1 + ¼).

 

80 / 2   = 40     remainder 0
40/2     = 20     remainder 0

20 / 2   = 10     remainder 0

10 / 2   = 5       remainder 0

5 / 2     = 2       remainder 1

2 / 2     = 1       remainder 0

1 / 2     = 1       remainder 1

 

Thus decimal 80 = 1010000 binary and decimal 80.09375 = 1010000.00011 binary.  To get
the binary point to be after the first 1, we move it six places to the left, so the normalized
form of the number is 1.01000000011 · 26, as expected.  For convenience, we write this as
1.0100 0000 0110 · 26.

 


Extended Example: Avogadro’s Number.

Up to this point, we have discussed the normalized representation of positive real numbers
where the conversion from decimal to binary can be done exactly for both the integer and
fractional parts.  We now consider conversion of very large real numbers in which it is not
practical to represent the integer part, much less convert it to binary.

 

We now discuss a rather large floating point number: 6.023 · 1023.  This is Avogadro’s
number.  We shall convert this to normalized form and use the opportunity to discuss a
number of issues associated with floating point numbers in general.

 

Avogadro’s number arises in the study of chemistry.  This number relates the atomic weight
of an element to the number of atoms in that many grams of the element.  The atomic weight
of oxygen is 16.00, as a result of which there are about 6.023 · 1023 atoms in 16 grams of
oxygen.  For our discussion we use a more accurate value of 6.022142 · 1023 obtained from
the web sit of the National Institute of Standards (www.nist.gov). 

 

We first remark that the number is determined by experiment, so it is not known exactly.  We
thus see one of the main scientific uses of this notation – to indicate the precision with which
the number is known.  The above should be read as (6.022142 ± 0.0000005) · 1023, that is to
say that the best estimate of the value is between 6.0221415 · 1023 and 6.0221425 · 1023, or
between 602, 214, 150, 000, 000, 000, 000, 000 and 602, 214, 250, 000, 000, 000, 000, 000. 
Here we see another use of scientific notation – not having to write all these zeroes.

 

Again, we use logarithms and anti-logarithms to convert this number to a power of two.  The
first question is how accurately to state the logarithm.  The answer comes by observing that
the number we are converting is known to seven digit’s precision.  Thus, the most accuracy
that makes sense in the logarithm is also seven digits.

 

In base-10 logarithms log(6.022142 · 1023) = 23.0 + log(6.022142).  To seven digits, this last
number is 0.7797510, so log(6.022142 · 1023) = 23.7797510.

 

We now use the fact that log(2.0) = 0.3010300 to seven decimal places to solve the equation
            2X = (100.3010300)X = 10.23.7797520 or 0.30103·X = 23.7797510 for X = 78.9946218.

 

If we use NA to denote Avogadro’s number, the first thing we have discovered from this
tedious analysis is that 278 < NA < 279, and that NA » 279.  The representation of the number in
normal form is thus of the form 1.f · 278, where the next step is to determine f.  To do this,
we obtain the decimal representation of 278.

 

Note that 278 = (100.30103)78 = 1023.48034 = 100.48034 · 1023 = 3.022317 · 1023.
But 6.022142 / 3.022317 = 1.992558, so NA = 1.992558 · 278, and f = 0.992558.

 


To complete this problem, we obtain the binary equivalent of 0.992558.

0.992558 · 2 = 1.985116        1
0.985116 · 2 = 1.970232        1
0.970232 · 2 = 1.949464        1
0.949464 · 2 = 1.880928        1
0.880928 · 2 = 1.761856        1
0.761856 · 2 = 1.523712        1
0.523712 · 2 = 1.047424        1

0.047424 · 2 = 0.094848        0
0.094848 · 2 = 0.189696        0
0.189696 · 2 = 0.379392        0
0.379392 · 2 = 0.758784        0

0.758784 · 2 = 1.517568        1

 

The desired form is 1.1111 1110 0001 · 278.

 

IEEE Standard 754 Floating Point Numbers

There are two primary formats in the IEEE 754 standard; single precision and double
precision.  We shall study the single precision format.

 

The single precision format is a 32–bit format.  From left to right, we have
            1 sign bit; 1 for negative and 0 for non-negative
            8 exponent bits
            23 bits for the fractional part of the mantissa.

 

The eight-bit exponent field stores the exponent of 2 in excess-127 form, with the
exception of two special bit patterns.

            0000 0000       numbers with these exponents are denormalized
            1111 1111       numbers with these exponents are infinity or Not A Number

 

Before presenting examples of the IEEE 754 standard, we shall examine the concept of NaN
or Not a Number.  In this discussion, we use some very imprecise terminology.

Consider the quotient 1/0.  The equation 1 / 0 = X is equivalent to solving for a number X
such that 0 · X = 1.  There is no such number X such that the result of multiplying it by 0 yields a 1. 
Loosely speaking, we say 1 / 0 = ¥; precisely speaking we use the notations of
the calculus and speak of limits to the fraction 1 / Y as the number Y approaches 0. 

Now consider the quotient 0/0.  Again we are asking for the number X such that 0 · X = 0. 
The difference here is that this equation is true for every number X.  In terms of the IEEE
standard, 0 / 0 is Not a Number, or NaN.

The number NaN can also be used for arithmetic operations that have no solutions, such as
taking the square root of –1 while limited to the real number system.  While this result
cannot be represented, it is definitely neither +¥ nor –¥ .


We now illustrate the standard by doing some conversions.

For the first example, consider the number –0.75.

 

To represent the number in the IEEE standard, first note that it is negative so that the sign bit
is 1.  Having noted this, we convert the number 0.75.

      0.75  · 2    = 1.5                1

      0.5    · 2    = 1.0                1

 

Thus, the binary equivalent of decimal 0.75 is 0.11 binary.  We must now convert this into
the normalized form 1.10 · 2–1.  Thus we have the key elements required.

            The power of 2 is –1, stored in Excess-127 as 126 = 0111 1110 binary.

            The fractional part is 10, possibly best written as 10000

 

Recalling that the sign bit is 1, we form the number as follows:

 

     1 0111 1110 10000

 

We now group the binary bits by fours from the left until we get only 0’s.

 

     1011 1111 0100 0000

 

Since trailing zeroes are not significant in fractions, this is equivalent to

 

     1011 1111 0100 0000 0000 0000 0000 0000

 

or BF40 0000 in hexadecimal.

 

As another example, we revisit a number converted earlier.  We have shown that

80.09375 = 1.0100 0000 0110 · 26.  This is a positive number, so the sign bit is 0.  As an
Excess–127 number, 6 is stored as 6 + 127 = 133 = 1000 0101 binary.  The fractional part
of the number is 0100 0000 0110 0000, so the IEEE representation is

     0 1000 0101 0100 0000 0110 0000

 

Regrouping by fours from the left, we get the following

 

     0100 0010 1010 0000 0011 0000

In hexadecimal this number is 42A030, or 42A0 3000 as an eight digit hexadecimal
number.  Note that all single–precision values should be represented with 8 hex digits.

Here is the general rule for floating–point numbers:
            single precision            32 bit values                represent with 8 hexadecimal digits
            double precision          64 bit values                represent with 16 hex digits.


Some Examples “In Reverse”

We now consider another view on the IEEE floating point standard – the “reverse” view. 
We are given a 32-bit number, preferably in hexadecimal form, and asked to produce the
floating-point number that this hexadecimal string represents.  As always, in interpreting any
string of binary characters, we must be told what standard to apply – here the IEEE-754 single precision standard.

 

First, convert the following 32-bit word, represented by eight hexadecimal digits, to the
floating-point number being represented.

 

            0000 0000              // Eight hexadecimal zeroes representing 32 binary zeroes

 

The answer is 0.0.  This is a result that should be memorized.

 

 

The question in the following paragraph was taken from a mid–term exam for an
introductory course in computer organization.  The paragraphs following were taken
from the answer key for that exam.

 

Give the value of the real number (in standard decimal representation) represented by
the following 32-bit words stored as an IEEE standard single precision.
            a)            4068 0000
            b)            42E8 0000
            c)            C2E8 0000
            d)            C380 0000

            e)            C5FC 0000

The first step in solving these problems is to convert the hexadecimal to binary.

a)   4068 0000 = 0100 0000 0110 1000 0000 0000 0000 0000
Regroup to get 0  1000 0000  1101 0000 etc. (trailing zeroes can be ignored)

Thus    s = 0 (not a negative number)
            p + 127 = 100000002 = 12810, so p = 1
and                     m = 1101, so 1.m = 1.1101 and the number is 1.1101·21 = 11.1012.

But 11.1012 = 2 + 1 + 1/2 + 1/8 = 3 + 5/8 = 3.625.

 

 

b)   42E8 0000 = 0100 0010 1110 1000 0000 0000 0000 0000
Regroup to get 0  1000 0101  1101 0000 etc
Thus    s = 0 (not a negative number)
            p + 127 = 100001012 = 128 + 4 + 1 = 133, hence p = 6
and                     m = 1101, so 1.m = 1.1101 and the number is 1.1101·26 = 11101002
But 11101002 = 64 + 32 + 16 + 4 = 96 + 20 = 116 = 116.0

 


c)   C2E80 0000 = 1100 0010 1110 1000 0000 0000 0000 0000
Regroup to get 1 1000 0101 1101 0000 etc.
Thus    s = 1 (a negative number) and the rest is the same as b).  So – 116.0

 

 

d)   C380 0000 = 1100 0011 1000 0000 0000 0000 0000 0000

Regroup to get 1  1000 0111  0000 0000 0000 0000 0000 000

Thus    s = 1 (a negative number)
            p + 127 = 100001112 = 128 + 7 = 135; hence p = 8.
            m = 0000, so 1.m = 1.0 and the number is  – 1.0 · 28 =  – 256.0

 

 

e)   C5FC 0000 = 1100 0101 1111 1100 0000 0000 0000 0000
Regroup to get 1  1000 1011  1111 1000 0000 0000 0000 000
Thus    s = 1 (a negative number)
            p + 127 = 1000 10112 = 128 + 8 + 2 + 1 = 139, so p = 12
            m = 1111 1000, so 1.m = 1.1111 1000

 

There are three ways to get the magnitude of this number.  The magnitude can be written
in normalized form as 1.1111 1000 · 212 = 1.1111 1000 · 4096, as 212 = 4096.

 

Method 1

If we solve this the way we have, we have to place four extra zeroes after the decimal point
to get the required 12, so that we can shift the decimal point right 12 places.

 

      1.1111 1000 · 212  = 1.1111 1000 0000 · 212 = 1 1111 1000 00002
                                    = 212 + 211+ 210 + 29 + 28 + 27
                                    = 4096 + 2048 + 1024 + 512 + 256 + 128 = 8064.

Method 2

We shift the decimal place only 5 places to the right (reducing the exponent by 5) to get
      1.1111 1000 · 212  = 1 1111 1.0 · 27
                                    = (25 + 24 + 23 + 22 + 21 + 20) · 27
                                    = (32 + 16 + 8 + 4 + 2 + 1) · 128 = 63 · 128 = 8064.

Method 3
This is an offbeat method, not much favored by students.

      1.1111 1000 · 212  = (1 + 2–1 + 2–2+ 2–3+ 2–4+ 2–5) · 212
                                    = 212 + 211+ 210 + 29 + 28 + 27
                                    = 4096 + 2048 + 1024 + 512 + 256 + 128 = 8064.

Method 4
This is another offbeat method, not much favored by students.

      1.1111 1000 · 212  = (1 + 2–1 + 2–2+ 2–3+ 2–4+ 2–5) · 212
                                    = (1 + 0.5 + 0.25 + 0.125 + 0.0625+ 0.03125) · 4096
                                    = 1.96875 · 4096 = 8064.

 

The answer is – 8064.0.


As a final example, we consider the IEEE standard representation of Avogadro’s number.
We have seen that NA » 1.1111 1110 0001 · 278.  This is a positive number; the sign bit is 0.

 

We now consider the representation of the exponent 78.  Now 78 + 127 = 205, so the
Excess-127 representation of 78 is 205 = 128 + 77 = 128 + 64 + 13 = 128 + 64 + 8 + 4 + 1. 
As an 8-bit binary number this is 1100 1101.  We already have the fractional part, so we get

 

     0 1100 1101 1111 1110 0001 0000

 

Grouped by fours from the left we get

 

     0110 0110 1111 1111 0000 1000 0000 0000,
or 66FF 0800 in hexadecimal.

 

Range and Precision

We now consider the range and precision associated with the IEEE single precision standard
using normalized numbers..  The range refers to the smallest and largest positive numbers
that can be stored.  Recalling that zero is not a positive number, we derive the smallest and
largest representable numbers.

 

In the binary the smallest normalized number is 1.0 · 2–126 and the largest number is a bit
less than 2.0 · 2127 = 2128.  Again, we use logarithms to evaluate these numbers.
            – 126 · 0.30103 = – 37.93 = – 38.0 + 0.07, so 2–126 = 1.07 · 10 -38, approximately.

            128 · 0.30103 = 38.53, so 2128 = 3.5 · 10 38, as 100.53 is a bit bigger than 3.2.

 

We now consider the precision associated with the standard.  Consider the decimal notation
1.23.  The precision associated with this is ± 0.005 as the number really represents a value
between 1.225 and 1.235 respectively.

 

The IEEE standard has a 23-bit fraction.  Thus, the precision associated with the standard is
1 part in 224 or 1 part in 16 · 220 = 16 · 1048576 = 16777216.  This accuracy is more precise
than 1 part in 107, or seven digit precision.

 

Denormalized Numbers

We shall see in a bit that the range of normalized numbers is approximately 10 –38 to 10 38. 
We now consider what we might do with a problem such as the quotient 10 –20 / 10 30.  In
plain algebra, the answer is simply 10 –50, a very small positive number.  But this number is
smaller than allowed by the standard.  We have two options for representing the quotient,
either 0.0 or some strange number that clearly indicates the underflow.  This is the purpose of
denormalized numbers – to show that the result of an operation is positive but too small
to be represented in standard format. 

 


Why Excess–127 Notation for the Exponent?

We have introduced two methods to be used for storing signed integers – two’s-complement
notation and excess–127 notation.  One might well ask why two’s-complement notation is
used to store signed integers while the excess–127 method is used for exponents in the
floating point notation.

The answer for integer notation is simple.  It is much easier to build an adder for integers
stored in two’s-complement form than it is to build an adder for integers in the excess
notation.  In the next chapter we shall investigate a two’s-complement adder.

So, why use excess–127 notation for the exponent in the floating point representation?  The
answer is best given by example.  Consider some of the numbers we have used as examples.

0011 1111 0100 0000 0000 0000 0000 0000 for 0.75

0100 0010 1010 0000 0011 0000 0000 0000 for 80.09375

0110 0110 1111 1111 0000 1000 0000 0000 for Avagadro’s number.

It turns out that the excess–127 notation allows the use of the integer compare unit to
compare floating point numbers.  Consider two floating point numbers X and Y.  Pretend that
they are integers and compare their bit patterns as integer bit patterns.  It viewed as an
integer, X is less than Y, then the floating point number X is less than the floating point Y. 
Note that we are not converting the numbers to integer form, just looking at the bit patterns
and pretending that they are integers.

Were the exponents stored in two’s–complement notation, we would have
0111 1111 1100 0000 0000 0000 0000 0000 for 0.75

0000 0011 0010 0000 0011 0000 0000 0000 for 80.09375

Comparing these as if they were integers makes the value 0.75 appear to be larger.  Recall
that it is the leftmost bit that is the sign bit in any standard numeric representation.
Pretending that each of these values is an integer makes both positive numbers.

Floating Point Equality: X == Y

Due to round off error, it is sometimes not advisable to check directly for equality of floating
point numbers.  A better method would be to use an acceptable relative error.  We borrow the
notation e from calculus to stand for a small number, and use the notation |Z| for the absolute
value of the number Z.

Here are two valid alternatives to the problematic statement (X == Y).

      1)   Absolute difference    |X – Y| £ e

      2)   Relative difference     |X – Y| £ e·(|X| +|Y|)

Note that this form of the second statement is preferable to computing the quotient
|X – Y| / (|X| +|Y|) which will be NaN (Not A Number) if X = 0.0 and Y = 0.0.

Bottom Line: In your coding with real numbers, decide what it means for two numbers to be
equal.  How close is close enough?  There are no general rules here, only cautions.  It is
interesting to note that one language (SPARK, a variant of the Ada programming language)
does not allow floating point comparison statements such as X == Y, but demands an
evaluation of the absolute value of the difference between X and Y.


The IBM Mainframe Floating–Point Formats

In this discussion, we shall adopt the bit numbering scheme used in the IBM documentation,
with the leftmost (sign) bit being number 0.  The IBM Mainframe supports three formats;
those representations with more bits can be seen to afford more precision.

      Single precision           32 bits       numbered 0 through 31,
      Double precision         64 bits       numbered 0 through 63, and
      Extended precision     128 bits     numbered 0 through 127.

As in the IEEE–754 standard, each floating point number in this standard is specified by
three fields: the sign bit, the exponent, and the fraction.  Unlike the IEEE–754 standard, the
IBM standard allocates the same number of bits for the exponent of each of its formats. 
The bit numbers for each of the fields are shown below.

Format

Sign bit

Bits for exponent

Bits for fraction

Single precision

0

1 – 7

8 – 31

Double precision

0

1 – 7

8 – 63

Extended precision

0

1 – 7

8 – 127

Note that each of the three formats uses eight bits to represent the exponent, in what is called
the characteristic field, and the sign bit.  These two fields together will be represented by
two hexadecimal digits in a one–byte field.

The size of the fraction field does depend on the format.
      Single precision           24 bits       6 hexadecimal digits,
      Double precision         56 bits       14 hexadecimal digits, and
      Extended precision     120 bits     30 hexadecimal digits.

The Characteristic Field

In IBM terminology, the field used to store the representation of the exponent is called the
“characteristic”.  This is a 7–bit field, used to store the exponent in excess–64 format; if
the exponent is E, then the value (E + 64) is stored as an unsigned 7–bit number.

Recalling that the range for integers stored in 7–bit unsigned format is 0 £ N £ 127, we have
0 £ (E + 64) £ 127, or –64 £ E £ 63.

Range for the Standard

We now consider the range and precision associated with the IBM floating point formats. 
The reader should remember that the range is identical for all of the three formats; only the
precision differs.  The range is usually specified as that for positive numbers, from a very
small positive number to a large positive number.  There is an equivalent range for negative
numbers.  Recall that 0 is not a positive number, so that it is not included in either range.

Given that the base of the exponent is 16, the range for these IBM formats is impressive.  It
is from somewhat less than 16–64 to a bit less than 1663.  Note that 1663 = (24)63 = 2252, and
16–64 = (24)–64 = 2–256 = 1.0 / (2256) and recall that log10(2) = 0.30103.  Using this, we compute
the maximum number storable at about (100.30103)252 = 1075.86 » 9·1075.  We may approximate
the smallest positive number at 1.0 / (36·1075) or about 3.0·10–77.  In summary, the following
real numbers can be represented in this standard: X = 0.0 and 3.0·10–77 < X < 9·1075.

One would not expect numbers outside of this range to appear in any realistic calculation.


Precision for the Standard

Unlike the range, which depends weakly on the format, the precision is very dependent on
the format used.  More specifically, the precision is a direct function of the number of bits
used for the fraction.  If the fraction uses F bits, the precision is 1 part in 2F.

We can summarize the precision for each format as follows.
      Single precision                  F = 24            1 part in 224.
      Double precision                 F = 56            1 part in 256.
      Extended precision           F = 120            1 part in 2120.

The first power of 2 is easily computed; we use logarithms to approximate the others.
            224                                                        = 16,777,216
            256        » (100.30103)56 = 1016.85              » 9·1016.
            2120      » (100.30103)120 = 1036.12 » 1.2·1036.

The argument for precision is quite simple.  Consider the single precision format, which is
more precise than 1 part in 10,000,000 and less precise than 1 part in 100,000,000.  In other
words it is better than 1 part in 107, but not as good as 1 in 108; hence we say 7 digits.

Range and Precision

We now summarize the range and precision for the three IBM Mainframe formats.

Format

Positive Range

Precision

Single Precision

3.0·10–77 < X < 9·1075

7 digits

Double Precision

3.0·10–77 < X < 9·1075

16 digits

Extended Precision

3.0·10–77 < X < 9·1075

36 digits

Representation of Floating Point Numbers

As with the case of integers, we shall most commonly use hexadecimal notation to represent
the values of floating–point numbers stored in the memory.  From this point, we shall focus
on the two more commonly used formats: Single Precision and Double Precision.

The single precision format uses a 32–bit number, represented by 8 hexadecimal digits.
The double precision format uses a 64–bit number, represented by 16 hexadecimal digits.

Due to the fact that the two formats use the same field length for the characteristic,
conversion between the two is quite simple.  To convert a single precision value to a double
precision value, just add eight hexadecimal zeroes.

Consider the positive number 128.0.
As a single precision number, the value is stored as      4280 0000.
As a double precision number, the value is stored as    4280 0000 0000 0000.

Conversions from double precision to single precision format will involve some rounding.
For example, consider the representation of the positive decimal number 123.45.  In a few
pages, we shall show that it is represented as follows.

As a double precision number, the value is stored as    427B 7333 3333 3333.
As a single precision number, the value is stored as      427B 7333.


The Sign Bit and Characteristic Field
We now discuss the first two hexadecimal digits in the representation of a floating–point
number in these two IBM formats.  In IBM nomenclature, the bits are allocated as follows.
            Bit 0                the sign bit
            Bits 1 – 7         the seven–bit number storing the characteristic.

Bit Number

0

1

2

3

4

5

6

7

Hex digit

0

1

Use

Sign bit

Characteristic (Exponent + 64)

Consider the four bits that comprise hexadecimal digit 0.  The sign bit in the floating–point
representation is the “8 bit” in that hexadecimal digit.  This leads to a simple rule.

If the number is not negative, bit 0 is 0, and hex digit 0 is one of 0, 1, 2, 3, 4, 5, 6, or 7.
If the number is negative,       bit 0 is 1, and hex digit 0 is one of 8, 9, A, B, C, D, E, or F.

Some Single Precision Examples
We now examine a number of examples, using the IBM single–precision floating–point
format.  The reader will note that the methods for conversion from decimal to hexadecimal
formats are somewhat informal, and should check previous notes for a more formal method. 
Note that the first step in each conversion is to represent the magnitude of the number in the
required form X·16E, after which we determine the sign and build the first two hex digits.

Example 1: Positive exponent and positive fraction.

The decimal number is 128.50.  The format demands a representation in the form X·16E,
with 0.625 £ X < 1.0.  As 128 £ X < 256, the number is converted to the form X·162.
Note that 128 = (1/2)·162 = (8/16)·162 , and 0.5 = (1/512)·162 = (8/4096)·162.
Hence, the value is 128.50 = (8/16 + 0/256 + 8/4096)·162; it is 162·0x0.808.

The exponent value is 2, so the characteristic value is either 66 or 0x42 = 100 0010. 
The first two hexadecimal digits in the eight digit representation are formed as follows.

Field

Sign

Characteristic

Value

0

1

0

0

0

0

1

0

Hex value

4

2

The fractional part comprises six hexadecimal digits, the first three of which are 808.
The number 128.50 is represented as 4280 8000.

Example 2: Positive exponent and negative fraction.

The decimal number is the negative number –128.50.  At this point, we would normally
convert the magnitude of the number to hexadecimal representation.  This number has the
same magnitude as the previous example, so we just copy the answer; it is 162·0x0.808.

We now build the first two hexadecimal digits, noting that the sign bit is 1.

Field

Sign

Characteristic

Value

1

1

0

0

0

0

1

0

Hex value

C

2

The number 128.50 is represented as C280 8000.
Note that we could have obtained this value just by adding 8 to the first hex digit.

Example 3: Negative exponent and positive fraction.

The decimal number is 0.375.  As a fraction, this is 3/8 = 6/16.  Put another way, it is
160·0.375 = 160·(6/16).  This is in the required format X·16E, with 0.625 £ X < 1.0.

The exponent value is 0, so the characteristic value is either 64 or 0x40 = 100 0000. 
The first two hexadecimal digits in the eight digit representation are formed as follows.

Field

Sign

Characteristic

Value

0

1

0

0

0

0

0

0

Hex value

4

0

The fractional part comprises six hexadecimal digits, the first of which is a 6.
The number 0.375 is represented in single precision as          4060 0000.
The number 0.375 is represented in double precision as         4060 0000 0000 0000.

Example 4: A Full Conversion
The number to be converted is 123.45.  As we have hinted, this is a non–terminator.

Convert the integer part.
123 / 16 = 7 with remainder 11                       this is hexadecimal digit B.
  7 / 16 = 0 with remainder  7                       this is hexadecimal digit 7.
Reading bottom to top, the integer part converts as 0x7B.

Convert the fractional part.
0.45 · 16 = 7.20         Extract the 7,
0.20 · 16 = 3.20         Extract the 3,
0.20 · 16 = 3.20         Extract the 3,
0.20 · 16 = 3.20         Extract the 3, and so on.

In the standard format, this number is 162·0x0.7B33333333…...

The exponent value is 2, so the characteristic value is either 66 or 0x42 = 100 0010. 
The first two hexadecimal digits in the eight digit representation are formed as follows.

Field

Sign

Characteristic

Value

0

1

0

0

0

0

1

0

Hex value

4

2

The number 123.45 is represented in single precision as        427B 3333.
The number 0.375 is represented in double precision as         427B 3333 3333 3333.

Example 5: One in “Reverse”
We are given the single precision representation of the number.  It is 4110 0000.
What is the value of the number stored?  We begin by examination of the first two hex digits.

Field

Sign

Characteristic

Value

0

1

0

0

0

0

0

1

Hex value

4

1

The sign bit is 0, so the number is positive.  The characteristic is 0x41, so the exponent is
1 and the value may be represented by X·161.  The fraction field is 100 000, so the value is
161·(1/16) = 1.0.

Packed Decimal Formats

While the IBM mainframe provides three floating–point formats, it also provides another
format for use in what are called “fixed point” calculations.  The term “fixed point” refers to
decimal numbers in which the decimal point takes a predictable place in the number; money
transactions in dollars and cents are a good and very important example of this.

 

Consider a ledger such as might be maintained by a commercial firm.  This contains credits
and debits, normally entered as money amounts with dollars and cents.  The amount that
might be printed as “$1234.56” could easily be stored as the integer 123456 if the program
automatically adjusted to provide the implicit decimal point.  This fact is the basis for the
Packed Decimal Format developed by IBM in response to its business customers.

 

One may well ask “Why not use floating point formats for financial transactions?”.  We
present a fairly realistic scenario to illustrate the problem with such a choice.  This example
is based on your author’s experience as a consultant to a bank in Rochester, NY.

 

It is a fact that banks loan each other money on an overnight basis; that is, the bank borrows
the money at 6:00 PM today and repays it at 6:00 AM tomorrow.  While this may seem a bit
strange to those of us who think in terms of 20–year mortgages, it is an important practice.

Overnight loans in the amount of one hundred million dollars are not uncommon.

 

Suppose that I am a bank officer, and that another bank wants to borrow $100,000,000
overnight.  I would like to make the loan, but do not have the cash on hand.  On the other
hand, I know a bank that will lend me the money at a smaller interest rate.  I can make the
loan and pocket the profit.

 

Suppose that the borrowing bank is willing to pay 8% per year on the borrowed amount. 
This corresponds to a payback of (1.08)1/730 = 1.0001054, which is $10,543 in interest.

 

Suppose that I have to borrow the money at 6% per annum.  This corresponds to my paying
at a rate of (1.06)1/730 = 1.0000798, which is a cost of $7,982 to me.  I make $2,561.

 

Consider these numbers as single–precision floating point format in the IBM Mainframe.

My original money amount is   $100,000,000

The interest I make is                         $10,543

My principal plus interest is      $100,010,500            Note the truncation due to precision.
The interest I pay is                              $7,982
What I get back is                     $100,002,000            Again, note the truncation.

 

The use of floating–point arithmetic has cost me $561 for an overnight transaction.  I do not
like that.  I do not like numbers that are rounded off; I want precise arithmetic.

 

Almost all banks and financial institutions demanded some sort of precise decimal
arithmetic; IBM’s answer was the Packed Decimal format.

 


BCD (Binary Coded Decimal)

The best way to introduce the Packed Decimal Data format is to first present an earlier
format for encoding decimal digits. This format is called BCD, for “Binary Coded Decimal”. 
As may be inferred from its name, it is a precursor to EBCDIC (Extended BCD Interchange
Code) in addition to heavily influencing the Packed Decimal Data format.

We shall introduce BCD and compare it to the 8–bit unsigned binary previously discussed for
storing unsigned integers in the range 0 through 255 inclusive.  While BCD doubtless had
encodings for negative numbers, we shall postpone signed notation to Packed Decimal.

The essential difference between BCD and 8–bit binary is that BCD encodes each decimal in
a separate 4–bit field (sometimes called “nibble” for half–byte).  This contrasts with the usual
binary notation in which it is the magnitude of the number (and not the number of digits) that
determines whether or not it can be represented in the format.

We begin with a table of the BCD codes for each of the ten decimal digits.  These codes are
given in both binary and hexadecimal.  It will be important for future discussions to note that
these encodings are actually hexadecimal digits; they just appear to be decimal digits.

Digit

‘0’

‘1’

‘2’

‘3’

‘4’

‘5’

‘6’

‘7’

‘8’

‘9’

Binary

0000

0001

0010

0011

0100

0101

0110

0111

1000

1001

Hexadecimal

0

1

2

3

4

5

6

7

8

9

To emphasize the difference between 8–bit unsigned binary and BCD, we shall examine a
selection of two–digit numbers and their encodings in each system.

Decimal Number

8–bit binary

BCD (Represented in binary)

BCD (hexadecimal)

5

0000 0101

0000 0101

05

13

0000 1101

0001 0011

13

17

0001 0001

0001 0111

17

23

0001 0111

0010 0011

23

31

0001 1111

0011 0001

31

64

0100 0000

0110 0100

64

89

0101 1001

1000 1001

89

96

0110 0000

1001 0110

96

As a hypothetical aside, consider the storage of BCD numbers on a byte–addressable
computer.  The smallest addressable unit would be an 8–bit byte.  As a result of this, all BCD
numbers would need to have an even number of digits, as to fill up an integral number of
bytes.  Our solution to the storage of integers with an odd number of digits is to recall that a
leading zero does not change the value of the integer.

In this hypothetical scheme of storage:
      1             would be stored as             01,
      22           would be stored as             22,
      333         would be stored as         0333,
      4444       would be stored as         4444,
      55555     would be stored as     055555,   and
      666666   would be stored as     666666.


Packed Decimal Data

The packed decimal format should be viewed as a generalization of the BCD format with the
specific goal of handling the fixed point arithmetic so common in financial transactions.  The
two extensions of the BCD format are as follows:
      1.      The provision of a sign “half byte” so that negative numbers can be handled.
      2.      The provision for variable length strings.

While the term “fixed point” is rarely used in computer literature these days, the format is
very common.  Consider any transaction denominated in dollars and cents.  The amount will
be represented as a real number with exactly two digits to the right of the decimal point; that
decimal point has a fixed position in the notation, hence the name “fixed point”.

The packed decimal format provides for a varying number of digits, one per half–byte,
followed by a half–byte denoting the sign of the number.  Because of the standard byte
addressability issues, the number of half–bytes in the representation must be an even number;
given the one half–byte reserved for the sign, this implies an odd number of digits.

In the BCD encodings, we use one hexadecimal digit to encode each of the decimal digits. 
This leaves the six higher–valued hexadecimal digits (A, B, C, D, E, and F) with no use; in
BCD these just do not encode any values.  In Packed Decimal, each of these will encode a
sign.  Here are the most common hexadecimal digits used to represent signs.

Binary

Hexadecimal

Sign

Comment

1100

C

+

The standard plus sign

1101

D

The standard minus sign

1111

F

+

A plus sign seen in converted EBCDIC

We now move to the IBM implementation of the packed decimal format.  This section breaks
with the tone previously established in this chapter – that of discussing a format in general
terms and only then discussing the IBM implementation.  The reason for this change is
simple; the IBM implementation of the packed decimal format is the only one used.

The Syntax of Packed Decimal Format

      1.      The length of a packed decimal number may be from 1 to 31 digits; the
               number being stored in memory as 1 to 16 bytes.

      2.      The rightmost half–byte of the number contains the sign indicator.  In constants
               defined by code, this is 0xC for positive numbers and 0xD for negative.

      3.      The remaining number of half–bytes (always an odd number) contain the
               hexadecimal encodings of the decimal digits in the number.

      4.      The rightmost byte in the memory representation of the number holds one
               digit and the sign half–byte.  All other bytes hold two digits.

      5.      The number zero is always represented as the two digits 0C, never 0D.

      6.      Any number with an even number of digits will be converted to an equivalent
               number with a prepended “0” prior to storage as packed decimal.

      7.      Although the format allows for storage of numbers with decimal points, neither
               the decimal point nor any indication of its position is stored.  As an example,
               each of 1234.5, 123.45, 12.345, and 1.2345 is stored as 12345C.


There are two common ways to generate numbers in packed decimal format, and quite a
variety of instructions to operate on data in this format.  We might discuss these in later
chapters.  For the present, we shall just show a few examples.

1.   Store the positive number 144 in packed decimal format.

Note that the number 144 has an odd number of digits.  The format just adds the half–byte
for non–negative numbers, generating the representation 144C.  This value is often written
as 14 4C, with the space used to emphasize the grouping of half–bytes by twos.

2.   Store the negative number –1023 in packed decimal format.

Note that the magnitude of the number (1023) has an even number of digits, so the format
will prepend a “0” to produce the equivalent number 01023, which has an odd number of
digits.  The value stored is 01023D, often written as 01 02 3D.

2.   Store the negative number –7 in packed decimal format.

Note that the magnitude of the number (7) has an odd number of digits, so the format
just adds the sign half–byte to generate the representation 7D.

4.   Store the positive number 123.456 in packed decimal format.

Note that the decimal point is not stored.  This is the same as the storage of the number
123456 (which has a decidedly different value).  This number has an even number of digits,
so that it is converted to the equivalent value 0123456 and stored as 01 23 45 6C.

5.   Store the positive number 1.23456 in packed decimal format.

Note that the decimal point is not stored.  This is the same as the storage of the number
123456 (which has a decidedly different value).  This number has an even number of digits,
so that it is converted to the equivalent value 0123456 and stored as 01 23 45 6C.

6.   Store the positive number 12345.6 in packed decimal format.

Note that the decimal point is not stored.  This is the same as the storage of the number
123456 (which has a decidedly different value).  This number has an even number of digits,
so that it is converted to the equivalent value 0123456 and stored as 01 23 45 6C.

7.   Store the number 0 in packed decimal form.

Note that 0 is neither positive nor negative.  IBM convention treats the zero as a positive
number, and always stores it as 0C.

8.   Store the number 12345678901234567890 in packed decimal form.

Note that very large numbers are easily stored in this format.  The number has 20 digits, an
even number, so it must first be converted to the equivalent 012345678901234567890.  It is
stored as 01 23 45 67 89 01 23 45 67 89 0C.

Comparison: Floating–Point and Packed Decimal

Here are a few obvious comments on the relative advantages of each format.

1.   Packed decimal format can provide great precision and range, more that is required
      for any conceivable financial transaction.  It does not suffer from round–off errors.

2.   The packed decimal format requires the code to track the decimal points explicitly.
      This is easily done for addition and subtraction, but harder for other operations.
      The floating–point format provides automatic management of the decimal point.


Character Codes: ASCII

We now consider the methods by which computers store character data.  There are three
character codes of interest: ASCII, EBCDIC, and Unicode.  The EBCDIC code is only used
by IBM in its mainframe computer.  The ASCII code is by far more popular, so we consider
it first and then consider Unicode, which can be viewed as a generalization of ASCII.

 

The figure below shows the ASCII code.  Only the first 128 characters (Codes 00 – 7F in
hexadecimal) are standard.  There are several interesting facts.

 

Last Digit \ First Digit

0

1

2

3

4

5

6

7

0

NUL

DLE

SP

0

@

P

`

p

1

SOH

DC1

!

1

A

Q

a

q

2

STX

DC2

2

B

R

b

r

3

ETX

DC3

#

3

C

S

c

s

4

EOT

DC4

$

4

D

T

d

t

5

ENQ

NAK

%

5

E

U

e

u

6

ACK

SYN

&

6

F

V

f

v

7

BEL

ETB

`

7

G

W

g

w

8

BS

CAN

(

8

H

X

h

x

9

HT

EM

)

9

I

Y

i

y

A

LF

SUB

*

:

J

Z

j

z

B

VT

ESC

+

;

K

[

k

{

C

FF

FS

< 

L

\

l

|

D

CR

GS

-

=

M

]

m

}

E

SO

RS

.

> 

N

^

n

~

F

SI

US

/

?

O

_

o

DEL

 

As ASCII is designed to be an 8–bit code, several manufacturers have defined extended code
sets, which make use of the codes 0x80 through 0xFF (128 through 255).  One of the more
popular was defined by IBM.  None are standard; most books ignore them.  So do we.

 

Let X be the ASCII code for a digit.  Then X – ‘0’ = X – 30 is the value of the digit.
For example ASCII(‘7’) = 37, with value 37 – 30 = 7.

 

Let X be an ASCII code.    If ASCII(‘A’) £ X £ ASCII(‘Z’) then X is an upper case letter.
                                             If ASCII(‘a’) £ X £ ASCII(‘z’) then X is an lower case letter.

                                             If ASCII(‘0’) £ X £ ASCII(‘9’) then X is a decimal digit.

 

Let X be an upper-case letter.  Then ASCII ( lower_case(X) ) = ASCII ( X ) + 32

Let X be a lower case letter.  Then ASCII ( UPPER_CASE(X) ) = ASCII (X ) – 32.
The expressions are ASCII ( X ) + 20 and ASCII (X ) – 20 in hexadecimal.

NOTE:     The first 32 character codes (decimal values 0 through 31) are control
                  characters used to manage a communication process and are not printed.  For
                  example the BEL (07) character would ring a bell on old teletype equipment.


Character Codes: EBCDIC

The Extended Binary Coded Decimal Interchange Code (EBCDIC) was developed by IBM
in the early 1960’s for use on its System/360 family of computers.  It evolved from an older
character set called BCDIC, hence its name.

EBCDIC code uses eight binary bits to encode a character set; it can encode 256 characters. 
The codes are binary numeric values, traditionally represented as two hexadecimal digits.

Character codes 0x00 through 0x3F and 0xFF represent control characters.
            0x0D   is the code for a carriage return; this moves the cursor back to the left margin.
            0x20    is used by the ED (Edit) instruction to represent a packed digit to be printed.
            0x21    is used by the ED (Edit) instruction to force significance.
                        All digits, including leading 0’s, from this position will be printed.
            0x25    is the code for a line feed; this moves the cursor down but not horizontally.
            0x2F    is the BELL code; it causes the terminal to emit a “beep”.

Character codes 0x40 through 0x7F represent punctuation characters.
            0x40    is the code for a space character: “ ”.
            0x4B   is the code for a decimal point: “.”.
            0x4E    is the code for a plus sign: “+”.
            0x50    is the code for an ampersand: “&”.
            0x5B   is the code for a dollar sign: “$”.
            0x5C   is the code for an asterisk: “*”.
            0x60    is the code for a minus sign: “–”.
            0x6B   is the code for a comma: “,”.
            0x6F    is the code for a question mark: “?”.
            0x7C   is the code for the commercial at sign: “@”.

Character codes 0x81 through 0xA9 represent the lower case Latin alphabet.
            0x81 through 0x89      represent the letters “a” through “i”,
            0x91 through 0x99      represent the letters “j” through “r”, and
            0xA2 through 0xA9    represent the letters “s” through “z”.

Character codes 0xC1 through 0xE9 represent the upper case Latin alphabet.
            0xC1 through 0xC9    represent the letters “A” through “I”,
            0xD1 through 0xD9    represent the letters “J” through “R”, and
            0xE2 through 0xE9     represent the letters “S” through “Z”.

Character codes 0xF0 through 0xF9 represent the digits “0” through “9”.

NOTES:

      1.   The control characters are mostly used for network data transmissions.
            The ones listed above appear frequently in user code for terminal I/O.

      2.   There are gaps in the codes for the alphabetical characters.
            This is due to the origins of the codes for the upper case alphabetic characters
            in the card codes used on the IBM–029 card punch.

      3.   One standard way to convert an EBCDIC digit to its numeric value
            is to subtract the hexadecimal number 0xF0 from the character code.


An Abbreviated Table: The Common EBCDIC

Code

Char.

Comment

Code

Char.

Comment

Code

Char.

Comment

 

 

 

80

 

 

C0

}

Right brace

 

 

 

81

a

 

C1

A

 

 

 

 

82

b

 

C2

B

 

 

 

 

83

c

 

C3

C

 

 

 

 

84

d

 

C4

D

 

 

 

 

85

e

 

C5

E

 

 

 

 

86

f

 

C6

F

 

 

 

 

87

g

 

C7

G

 

0C

FF

Form feed

88

h

 

C8

H

 

0D

CR

Return

89

i

 

C9

I

 

16

BS

Back space

90

 

 

D0

{

Left brace

25

LF

Line Feed

91

j

 

D1

J

 

27

ESC

Escape

92

k

 

D2

K

 

2F

BEL

Bell

93

l

 

D3

L

 

40

SP

Space

94

m

 

D4

M

 

4B

.

Decimal

95

n

 

D5

N

 

4C

< 

 

96

o

 

D6

O

 

4D

(

 

97

p

 

D7

P

 

4E

+

 

98

q

 

D8

Q

 

4F

|

Single Bar

99

r

 

D9

R

 

50

&

 

A0

 

 

E0

\

Back slash

5A

!

 

A1

~

Tilde

E1

 

 

5B

$

 

A2

s

 

E2

S

 

5C

*

 

A3

t

 

E3

T

 

5D

)

 

A4

u

 

E4

U

 

5E

;

 

A5

v

 

E5

V

 

5F

­

Not

A6

w

 

E6

W

 

60

Minus

A7

x

 

E7

X

 

61

/

Slash

A8

y

 

E8

Y

 

6A

¦

Dbl. Bar

A9

z

 

E9

Z

 

6B

,

Comma

B0

^

Carat

F0

0

 

6C

%

Percent

B1

 

 

F1

1

 

6D

_

Underscore

B2

 

 

F2

2

 

6E

> 

 

B3

 

 

F3

3

 

6F

?

 

B4

 

 

F4

4

 

79

Apostrophe

B5

 

 

F5

5

 

7A

:

Colon

B6

 

 

F6

6

 

7B

#

Sharp

B7

 

 

F7

7

 

7C

@

At Sign

B8

 

 

F8

8

 

7D

'

Apostrophe

B9

 

 

F9

9

 

7E

=

Equals

BA

[

Left Bracket

 

 

 

7F

"

Quote

BB

]

R. Bracket

 

 

 

 

It is worth noting that IBM seriously considered adoption of ASCII as its method for internal
storage of character data for the System/360.  The American Standard Code for Information
Interchange was approved in 1963 and supported by IBM.  However the ASCII code set was
not compatible with the BCDIC used on a very large installed base of support equipment,
such as the IBM 026.  Transition to an incompatible character set would have required any
adopter of the new IBM System/360 to also purchase or lease an entirely new set of
peripheral equipment; this would have been a deterrence to early adoption.

The figure below shows a standard 80–column IBM punch card produced by the IBM 029
card punch.  This shows the card punch codes used to represent some EBCDIC characters.

IBM_029PunchCodes

The structure of the EBCDIC, used for internal character storage on the System/360 and later
computers, was determined by the requirement for easy translation from punch card codes. 
The table below gives a comparison of the two coding schemes.

Character

Punch Code

EBCDIC

‘0’

0

F0

‘1’

1

F1

‘9’

9

F9

‘A’

12 – 1

C1

‘B’

12 – 2

C2

‘I’

12 – 9

C9

‘J’

11 – 1

D1

‘K’

11 – 2

D2

‘R’

11 – 9

D9

‘S’

0 – 2

E2

‘T’

0 – 8

E3

‘Z’

0 – 9

E9

Remember that the punch card codes
represent the card rows punched.  Each
digit was represented by a punch in a
single row; the row number was
identical to the value of the digit being
encoded.

The EBCDIC codes are eight–bit binary
numbers, almost always represented as
two hexadecimal digits.  Some IBM
documentation refers to these digits as:
      The first digit is the zone potion,
      The second digit is the numeric.

 

 


A comparison of the older card punch codes with the EBCDIC shows that its design was
intended to facilitate the translation.  For digits, the numeric punch row became the numeric
part of the EBCDIC representation, and the zone part was set to hexadecimal F.  For the
alphabetical characters, the second numeric row would become the numeric part and the
first punch row would determine the zone portion of the EBCDIC.

This matching with punched card codes explains the “gaps” found in the EBCDIC set. 
Remember that these codes are given as hexadecimal numbers, so that the code immediately
following C9 would be CA (as hexadecimal A is decimal 10).  But the code for ‘J’ is not
hexadecimal CA, but hexadecimal D1.  Also, note that the EBCDIC representation for the
letter ‘S’ is not E1 but E2.  This is a direct consequence of the design of the punch cards.

Character Codes: UNICODE

The UNICODE character set is a generalization of the ASCII character set to allow for the
fact that many languages in the world do not use the Latin alphabet.  The important thing to
note here is that UNICODE characters consume 16 bits (two bytes) while ASCII and
EBCDIC character codes are 8 bits (one byte) long.  This has some implications in
programming with modern languages, such as Visual Basic and Visual C++, especially in
allocation of memory space to hold strings.  This seems to be less of an issue in Java.

An obvious implication of the above is that, while each of ASCII and EBCDIC use two
hexadecimal digits to encode a character, UNICODE uses four hexadecimal digits.  In part,
UNICODE was designed as a replacement for the ad–hoc “code pages” then in use.  These
pages allowed arbitrary 256–character sets by a complete redefinition of ASCII, but were
limited to 256 characters.  Some languages, such as Chinese, require many more characters.

UNICODE is downward compatible with the ASCII code set; the characters represented by
the UNICODE codes 0x0000 through 0x007F are exactly those codes represented by the
standard ASCII codes 0x00 through 0x7F.  In other words, to convert standard ASCII to
correct UNICODE, just add two leading hexadecimal 0’s and make a two–byte code.

The origins of Unicode date back to 1987 when Joe Becker from Xerox and Lee Collins and
Mark Davis from Apple started investigating the practicalities of creating a universal
character set. In August of the following year Joe Becker published a draft proposal for an
"international/multilingual text character encoding system, tentatively called Unicode."
In this document, entitled Unicode 88, he outlined a 16 bit character model:

“Unicode is intended to address the need for a workable, reliable world text
encoding. Unicode could be roughly described as "wide-body ASCII" that has
been stretched to 16 bits to encompass the characters of all the world's living
languages. In a properly engineered design, 16 bits per character are more than
sufficient for this purpose.”

In fact the 16–bit (four hexadecimal digit) code scheme has proven not to be adequate to
encode every possible character set.  The original code space (0x0000 – 0xFFFF) was
defined as the “Basic Multilingual Plane”, or BMP.  Supplementary planes have been
added, so that as of September 2008 there were over 1,100,000 “code points” in UNICODE.


Here is a complete listing of the character sets and languages supported by the Basic
Multilingual Plane.  The source is http://www.ssec.wisc.edu/~tomw/java/unicode.html.

Range

Decimal

Name

0x0000-0x007F

0-127

Basic Latin (what we normally use)

0x0080-0x00FF

128-255

Latin-1 Supplement

0x0100-0x017F

256-383

Latin Extended-A

0x0180-0x024F

384-591

Latin Extended-B

0x0250-0x02AF

592-687

IPA Extensions

0x02B0-0x02FF

688-767

Spacing Modifier Letters

0x0300-0x036F

768-879

Combining Diacritical Marks

0x0370-0x03FF

880-1023

Greek

0x0400-0x04FF

1024-1279

Cyrillic

0x0530-0x058F

1328-1423

Armenian

0x0590-0x05FF

1424-1535

Hebrew

0x0600-0x06FF

1536-1791

Arabic

0x0700-0x074F

1792-1871

Syriac

0x0780-0x07BF

1920-1983

Thaana

0x0900-0x097F

2304-2431

Devanagari

0x0980-0x09FF

2432-2559

Bengali

0x0A00-0x0A7F

2560-2687

Gurmukhi

0x0A80-0x0AFF

2688-2815

Gujarati

0x0B00-0x0B7F

2816-2943

Oriya

0x0B80-0x0BFF

2944-3071

Tamil

0x0C00-0x0C7F

3072-3199

Telugu

0x0C80-0x0CFF

3200-3327

Kannada

0x0D00-0x0D7F

3328-3455

Malayalam

0x0D80-0x0DFF

3456-3583

Sinhala

0x0E00-0x0E7F

3584-3711

Thai

0x0E80-0x0EFF

3712-3839

Lao

0x0F00-0x0FFF

3840-4095

Tibetan

0x1000-0x109F

4096-4255

Myanmar

0x10A0-0x10FF

4256-4351

Georgian

0x1100-0x11FF

4352-4607

Hangul Jamo

0x1200-0x137F

4608-4991

Ethiopic

0x13A0-0x13FF

5024-5119

Cherokee

0x1400-0x167F

5120-5759

Unified Canadian Aboriginal Syllabics

0x1680-0x169F

5760-5791

Ogham

0x16A0-0x16FF

5792-5887

Runic

0x1780-0x17FF

6016-6143

Khmer

0x1800-0x18AF

6144-6319

Mongolian

Range

Decimal

Name

0x1E00-0x1EFF

7680-7935

Latin Extended Additional

0x1F00-0x1FFF

7936-8191

Greek Extended

0x2000-0x206F

8192-8303

General Punctuation

0x2070-0x209F

8304-8351

Superscripts and Subscripts

0x20A0-0x20CF

8352-8399

Currency Symbols

0x20D0-0x20FF

8400-8447

Combining Marks for Symbols

0x2100-0x214F

8448-8527

Letter–like Symbols

0x2150-0x218F

8528-8591

Number Forms

0x2190-0x21FF

8592-8703

Arrows

0x2200-0x22FF

8704-8959

Mathematical Operators

0x2300-0x23FF

8960-9215

Miscellaneous Technical

0x2400-0x243F

9216-9279

Control Pictures

0x2440-0x245F

9280-9311

Optical Character Recognition

0x2460-0x24FF

9312-9471

Enclosed Alphanumerics

0x2500-0x257F

9472-9599

Box Drawing

0x2580-0x259F

9600-9631

Block Elements

0x25A0-0x25FF

9632-9727

Geometric Shapes

0x2600-0x26FF

9728-9983

Miscellaneous Symbols

0x2700-0x27BF

9984-10175

Dingbats

0x2800-0x28FF

10240-10495

Braille Patterns

0x2E80-0x2EFF

11904-12031

CJK Radicals Supplement

0x2F00-0x2FDF

12032-12255

Kangxi Radicals

0x2FF0-0x2FFF

12272-12287

Ideographic Description Characters

0x3000-0x303F

12288-12351

CJK Symbols and Punctuation

0x3040-0x309F

12352-12447

Hiragana

0x30A0-0x30FF

12448-12543

Katakana

0x3100-0x312F

12544-12591

Bopomofo

0x3130-0x318F

12592-12687

Hangul Compatibility Jamo

0x3190-0x319F

12688-12703

Kanbun

0x31A0-0x31BF

12704-12735

Bopomofo Extended

0x3200-0x32FF

12800-13055

Enclosed CJK Letters and Months

0x3300-0x33FF

13056-13311

CJK Compatibility

0x3400-0x4DB5

13312-19893

CJK Unified Ideographs Extension A

0x4E00-0x9FFF

19968-40959

CJK Unified Ideographs

0xA000-0xA48F

40960-42127

Yi Syllables

0xA490-0xA4CF

42128-42191

Yi Radicals

0xAC00-0xD7A3

44032-55203

Hangul Syllables

0xD800-0xDB7F

55296-56191

High Surrogates

0xDB80-0xDBFF

56192-56319

High Private Use Surrogates

Range

Decimal

Name

0xDC00-0xDFFF

56320-57343

Low Surrogates

0xE000-0xF8FF

57344-63743

Private Use

0xF900-0xFAFF

63744-64255

CJK Compatibility Ideographs

0xFB00-0xFB4F

64256-64335

Alphabetic Presentation Forms

0xFB50-0xFDFF

64336-65023

Arabic Presentation Forms-A

0xFE20-0xFE2F

65056-65071

Combining Half Marks

0xFE30-0xFE4F

65072-65103

CJK Compatibility Forms

0xFE50-0xFE6F

65104-65135

Small Form Variants

0xFE70-0xFEFE

65136-65278

Arabic Presentation Forms-B

0xFEFF-0xFEFF

65279-65279

Specials

0xFF00-0xFFEF

65280-65519

Halfwidth and Fullwidth Forms

0xFFF0-0xFFFD

65520-65533

Specials

Here is a bit of the Greek alphabet as encoded in the BMP.


For those with more esoteric tastes, here is a small sample of Cuneiform in 32–bit Unicode.

Now we see some Egyptian hieroglyphics, also with the 32–bit Unicode encoding.

We close this chapter with a small sample of the 16–bit BMP encoding for the CJK
(Chinese, Japanese, & Korean) character set.  Unlike the above two examples (Cuneiform
and Egyptian hieroglyphics) this is a living language.

As a final note, we mention the fact that some fans of the Star Trek series have proposed that
the alphabet for the Klingon language be included in the Unicode 32–bit encodings.  So far,
they have inserted it in the Private Use section (0xE000-0xF8FF).  It is not yet recognized as
an official part of the Unicode standard.


 

Solved Problems
As is obvious from the variety of fonts, these problems have been assembled from a variety
of sources.  All of these problems have been used in a teaching context.

1.     What range of integers can be stored in an 16–bit word if

        a) the number is stored as an unsigned integer?

        b) the number is stored in two’s–complement form?

Answer:    a)   0 through 65,535 inclusive, or 0 through 216 – 1.
                  b) –32768 through 32767 inclusive, or –(215) through (215) – 1

2    You are given the 16–bit value, represented as four hexadecimal digits,
      and stored in two bytes.  The value is 0x812D.

      a)      What is the decimal value stored here, if interpreted as a packed decimal number?

      b)      What is the decimal value stored, if interpreted as a 16–bit two’s–complement
               integer?

      c)      What is the decimal value stored here, if interpreted as a 16–bit unsigned integer?

ANSWER:        The answers are found in the lectures for January 13 and January 20.

      a)      For a packed decimal number, the absolute value is 812 and the value is negative.
               The answer is –812.

      b)      To render this as a two’s–complement integer, one first has to convert to binary.
               Hexadecimal
812D converts to      1000 0001 0010 1101.  This is negative.
               Take the one’s complement to get  
0111 1110 1101 0010.
               Add 1 to get the positive value      
0111 1110 1101 0011.
               In hexadecimal, this is
7ED3, which converts to 7·163 + 14·162 + 13·16 + 3,
               or 7
·4096 + 14·256 + 13·16 + 3 = 28672 + 3584 + 208 + 3 = 32,467
               The answer is –32,467.

      c)      As an unsigned binary number the value is obtained by direct conversion from
               the hexadecimal value.  The value is 8
·163 + 1·162 + 2·16 + 13,
               or 8
·4096 + 1·256 + 2·16 + 13 = 32768 + 256 + 32 + 13 = 33069.

3.   Give the 8–bit two’s complement representation of the number – 98.

Answer:            98 = 96 + 2 = 64 + 32 + 2, so its binary representation is 0110 0010.
                          8–bit representation of + 98           0110 0010
                          One’s complement                           1001 1101
                          Add 1 to get                                     1001 1110       9E.

4.   Give the 16–bit two’s complement representation of the number – 98.

Answer:            The 8–bit representation of – 98 is      1001 1110
                          Sign extend to 16 bits          1111 1111 1001 1110


 

5    Convert the following decimal numbers to binary.

            a)  37.375        b)  93.40625

ANSWER:  Recall that the integer part and fractional part are converted separately.

      a)  37.375
            37 / 2                    = 18            rem 1
            18 / 2                     =  9            rem 0
            9 / 2                       =  4            rem 1
            4 / 2                        = 2            rem 0
            2 / 2                        = 1            rem 0
            1 / 2                        = 0            rem 1   Answer: 100101.

            0.375 · 2           = 0.75
            0.75
· 2            =  1.50
            0.50
· 2             = 1.00
            0.00
· 2             = 0.00                        Answer 0.011              100101.011

      b)   93.40625
            93 / 2                    = 46            rem 1
            46 / 2                    = 23            rem 0
            23 / 2                    = 11            rem 1
            11 / 2                     =  5            rem 1
            5 / 2                       =  2            rem 1
            2 / 2                       =  1            rem 0
            1 / 2                       =  0            rem 1   Answer: 1011101

                        0.40625 · 2     = 0.8125
                        0.8125
· 2       = 1.6250
                        0.625
· 2         = 1.2500
                        0.25
· 2           = 0.5000
                        0.50
· 2           = 1.0000
                        0.00
· 2           = 0.0000                      Answer: 0.01101         1011101.01101

6    Convert the following hexadecimal number to decimal numbers.
      The numbers are unsigned.  Use as many digits as necessary
      a) 0x022    b) 0x0BAD     c) 0x0EF

ANSWER: A = 10, B = 11, C = 12, D = 13, E = 14, F = 15
      160 = 1,  161 = 16,  162 = 256,  163 = 4096,  164 = 65536
      a)   0x022  = 2
·16 + 2 = 32 + 2    = 34     34
      b)   0x0BAD         = 11
·162 + 10·16 + 13            = 11·256 + 10·16 + 13
            = 2816 + 160 + 13      = 2989 2989
      c)   0x0EF = 14
·16 + 15   = 224 + 15 = 239        239


 

7    Show the IEEE–754 single precision representation of the following real numbers.
      Show all eight hexadecimal digits associated with each representation.
      a)   0.0       b) – 1.0            c) 7.625           d) – 8.75

ANSWER:
a)   0.0 = 0x0000 0000                                                                                       0x0000 0000

b)   – 1.0                this is negative, so the sign bit is S = 1
      1.0 = 1.0
·20
      1.M = 1.0        so M = 0000
      P = 0                so P + 127 = 127   = 0111 11112

      Concatenate S | (P + 127) | M       1 0111 1111 0000
      Group by 4’s from the left           
1011 1111 1000 0
      Pad out the last to four bits         
1011 1111 1000 0000
      Convert to hex digits                    BF80
      Pad out to eight hexadecimal digits                                                           0xBF80 0000

c)   7.625               this is non-negative, so the sign bit is S = 0
      Convert 7.625 to binary.
                              7 = 4 + 2 + 1                                       01112
                              0.625 = 5/8 = 1/2 + 1/8                       .1012
                              7.625                                                   111.1012
      Normalize by moving the binary point
      two places to the left.                                                 1.11101
·22
      Thus saying that 22
£ 7.625 < 23.

      1.M = 1.11101   so M = 11101
      P = 2                so P + 127 = 129   = 1000 0001

      Concatenate S | (P + 127) | M       0 1000 0001 11101
      Group by 4’s from the left           
0100 0000 1111 01
      Pad out the last to four bits         
0100 0000 1111 0100
      Covert to hex digits                      40F4                                                      0x40F4 0000

d)   – 8.75              this is negative, so S = 1
      8.75 = 8 + 1/2 + 1/4                      1000.11
                                                            1.00011
·23

      1.M = 1.00011  so M = 00011
      P = 3                so P + 127 = 130   1000 0010
      Concatenate S | (P + 127) | M       1 1000 0010 00011
      Group by 4’s from the left            1100 0001 0000 11
      Pad out the last to four bits          1100 0001 0000 1100
      Convert to hex digits                    C10C                                                    0xC10C 0000

 


 

8    Give the value of the real number (in standard decimal representation) represented by the
following 32-bit words stored as an IEEE standard single precision.
      a)   4068 0000       b)         42E8 0000       c)         C2E8 0000

ANSWERS:

The first step in solving these problems is to convert the hexadecimal to binary.

a)   4068 0000 = 0100 0000 0110 1000 0000 0000 0000 0000
Regroup to get 0  1000 0000  1101 0000 etc.

Thus    s = 0 (not a negative number)
            p + 127 = 100000002 = 12810, so p = 1
and                  m = 1101, so 1.m = 1.1101 and the number is 1.1101
·21 = 11.1012.

But 11.1012 = 2 + 1 + 1/2 + 1/8 = 3 + 5/8 = 3.625.

b)   42E8 0000 = 0100 0010 1110 1000 0000 0000 0000 0000
Regroup to get 0  1000 0101  1101 0000 etc
Thus    s = 0 (not a negative number)
            p + 127 = 100001012 = 128 + 4 + 1 = 133, hence p = 6
and                  m = 1101, so 1.m = 1.1101 and the number is 1.1101
·26 = 11101002
But 11101002 = 64 + 32 + 16 + 4 = 96 + 20 = 116 = 116.0

c)   C2E80 0000 = 1100 0010 1110 1000 0000 0000 0000 0000
Regroup to get 1 1000 0101 1101 0000 etc.
Thus    s = 1 (a negative number) and the rest is the same as b).  So – 116.0

9    Consider the string of digits “2108”.
      a)   Show the coding of this digit string in EBCDIC.
      b)   How many bytes does this encoding take?

ANSWER:  F2 F1 F0 F8.  Four bytes.

10  Consider the positive number 2108.
      a)   Show the representation of this number in packed decimal format.
      b)   How many bytes does this representation take?

ANSWER:  To get an odd number of decimal digits, this must be represented
                     with a leading 0, as 02108.  02 10 8C.  Three bytes.

11  The following block of bytes contains EBCDIC characters.
      Give the English sentence represented.

E3 C8 C5 40 C5 D5 C4 4B

Answer:             E3 C8 C5 40 C5 D5 C4 4B
           T  H  E     E  N  D  .  “THE END.”


 

12  Perform the following sums assuming that each is a hexadecimal number.
                           Show the results as 16–bit (four hexadecimal digit) results.

      a)   123C + 888C

      b)   123C + 99D

ANSWER:  a)   In hexadecimal       C + C = 18                  (12 + 12 = 24 = 16 + 8).
                                                            3 + 8 + 1 = C              (Decimal 12 is 0xC)
                                                            2 + 8 + 0 = A              (Decimal 10 is 0xA)
                                                            1 + 8 + 0 = 9.              9AC8

                     b)   In hexadecimal       C + D = 19                  (12 + 13 = 25 = 16 + 9)
                                                            3 + 9 + 1 = D              (Decimal 13 is 0xD)
                                                            2 + 9 + 0 = B               (Decimal 11 is 0xB)
                                                            1 + 0 + 0 = 1               1BD9

Each of the previous two problems uses numbers written as 123C and 888C.

The numbers in the problem on packed decimal are not the same as those in
the problem on hexadecimal.  They just look the same.

Interpreted as packed decimal:           123C is interpreted as the positive number 123, and
                                                            888C is interpreted as the positive number 888.

If we further specify that each of these is to be read as an integer value, then we have
the two numbers +123 and +888.

Interpreted as hexadecimal, 123C is interpreted as the decimal number
1
·163 + 2·162 + 3·16 + 12 = 4096 + 512 + 48 + 12 = 4,668

Interpreted as hexadecimal, 888C is interpreted as the decimal number
8
·163 + 8·162 + 8·16 + 12 = 32768 + 2048 + 128 + 12 = 34,968

As for the number 1011, it translates to 0x3F3.

13  The two–byte entry shown below can be interpreted in a number of ways.

VALUE     DC X‘021D’

      a)      What is its decimal value if it is interpreted as an unsigned binary integer?
      b)      What is its decimal value if it is interpreted as a packed decimal value?

ANSWER: As a binary integer, its value is 2·162 + 1·16 + 13 = 512 + 16 + 13 = 541.

                     As a packed decimal, this has value – 21.

 


 

14  A given computer uses byte addressing with the little-endian structure.
      The following is a memory map, with all values expressed in hexadecimal.

 

Address

104

105

106

107

108

109

10A

10B

10C

10D

Value

C2

3F

84

00

00

00

9C

C1

C8

C0

 

      What is the value (as a decimal real number; e.g. 203.75 ) of the floating point
      number stored at address 108?  Assume IEEE–754 single precision format.

 

Answer:  The first thing to notice is that the memory is byte-addressable.  That means that
each address takes holds one byte or eight bits.  The IEEE format for single precision
numbers calls for 32 bits to be stored, so the number takes four bytes of memory.

 

In byte addressable systems, the 32-bit entry at address 108 is stored in the four bytes
with addresses 108, 109, 10A, and 10B.  The contents of these are 00, 00, 9C, and C1.

 

The next thing to do is to get the four bytes of the 32-bit number in order.

This is a little-endian memory organization, which means that the LSB is stored at address
108 and the MSB is stored as address 10B.  In correct order, the four bytes are
            C1  9C  00  00.  In binary, this becomes
            1100 0001   1001 1100   0000 0000   0000 0000.  Breaking into fields we get
            1100 0001 1001 1100 0000 0000 0000 0000, with the exponent field in bold, or

            1  1000 0011  0011 1000.

      Thus             s = 1             a negative number

                           e + 127 = 1000 00112 = 128 + 2 + 1 = 131, so e = 4
                           m = 00111

      So the number’s magnitude is 1.001112 · 24 = 10011.12 = 16 + 2 + 1 + 0.5 = 19.5,

      and the answer is – 19.5.

15  Consider the 32-bit number represented by the eight hexadecimal digits
      BEEB 0000.  What is the value of the floating point number represented by this
      bit pattern assuming that the IEEE-754 single-precision standard is used?

 

ANSWER:  First recall the binary equivalents: B = 1011, E = 1110, and 0 = 0000.

Convert the hexadecimal string to binary

 

Hexadecimal:  B    E    E    B    0    0    0    0

Binary:     1011 1110 1110 1011 0000 0000 0000 0000

 

Regroup the binary according to the 1 | 8 | 23 split required by the format.

 

Binary:     1011 1110 1110 1011 0000 0000 0000 0000
Split:      1 011 1110 1 110 1011 0000 0000 0000 0000
Regrouped:  1  0111 1101  1101 0110 0000 0000 0000 000

 

The fields in the expression are now analyzed.

Sign bit:              S = 1             this will become a negative number

Exponent:

The field contains 0111 1101, or 64 + 32 + 16 + 8 + 4 + 1 = 96 + 24 + 5 = 125.  This number
may be more easily derived by noting that 0111 1111 = 127 and this is 2 less.

The exponent is given by P + 127 = 125, or P = – 2.  The absolute value of the number being
represented should be in the range [0.25, 0.50), or 0.25 £ N  < 0.50.

Mantissa:

The mantissa field is 1101 0110, so 1.M = 1.1101 0110.

In decimal, this equals 1 + 1/2 + 1/4 + 1/16 + 1/64 + 1/128, also written as
                                    (128 + 64 + 32 + 8 + 2 + 1) / 128 = (192 + 40 + 3) / 128 = 235 / 128.

The magnitude of the number equals 235 / 128 · 1/4 = 235 / 512 = 0.458984375.

16  The following are two examples of the hexadecimal representation of
              floating–point numbers stored in the IBM single–precision format.
              Give the decimal representation of each.  Fractions (e.g., 1/8) are acceptable.

      a)     C1 64 00 00

      b)    3F 50 00 00

ANSWER:  a)   First look at the sign and exponent byte.  This is 0xC1, or 1100 0001.

Bit

0

1

2

3

4

5

6

7

Value

1

1

0

0

0

0

0

1

The sign bit is 1, so this is a negative number.
Stripping the sign bit, the exponent field is 0100 0001 or 0x41 = decimal 65.
We have (Exponent + 64) = 65, so the exponent is 1.

The value is 161·F, where F is 0x64, or 6/16 + 4/256.
The magnitude of the number is 161
·(6/16 + 4/256) = 6 + 4/16 = 6.25.  The value is –6.25.

                     b)   First look at the sign and exponent byte.  This 0x3F, or 0011 1111

The sign bit is 0, so this is a non–negative number.
The exponent field is 0x3F = 3
·16 + 15 = decimal 63 = 64 – 1.

The value is 16–1·F, where F is 0x50, or 5/16.
The magnitude of the number is 16–1
·(5/16) = 5/256 = 0.01953125.

17  These questions refer to the IBM Packed Decimal Format.

      a)      How many bytes are required to represent a 3–digit integer?

      b)      Give the Packed Decimal representation of the positive integer 123.

      c)      Give the Packed Decimal representation of the negative integer –107.

ANSWER:        Recall that each decimal digit is stored as a hexadecimal digit, and
                           that the form calls for one hexadecimal digit to represent the sign.

      a)      One needs four hexadecimal digits, or two bytes, to represent three decimal digits.

      b)                  12 3C                     c)         10 7D

 


 

18  These questions also refer to the IBM Packed Decimal Format.

      a)      How many decimal digits can be represented in Packed Decimal form
               if three bytes (8 bits each) are allocated to store the number?

      b)      What is the Packed Decimal representation of the largest integer stored in 3 bytes?

ANSWER:        Recall that N bytes will store 2·N hexadecimal digits.  With one of these
                           reserved for the sign, this is (2
·N – 1) decimal digits.

      a)      3 bytes can store five decimal digits.

      b)      The largest integer is 99,999.  It is represented as 99 99 9C.

19  Convert the following numbers to their representation IBM Single Precision
      floating point and give the answers as hexadecimal digits.

      a)      123.75

      b)      –123.75

ANSWER:
      a)     
First convert the number to hexadecimal.

The whole number conversion:           123 / 16 = 7 with remainder = 11 (B)
                                                            7 / 16 = 0 with remainder 7.   123 = 7B.

The fractional part conversion:           .75·16 = 12 (C).  The number is 7B.C

The number can be represented as 162 · 0.7BC; the exponent is 2.

The exponent stored with excess 64, thus it is 66 or X‘42’.

Appending the fractional part, we get X‘427BC’.

Add three hexadecimal zeroes to pad out the answer to X‘427B C000

b)   The only change here is to add the sign bit as the leftmost bit.

In the positive number, the leftmost byte was X‘42’, which in binary would be

Bit

0

1

2

3

4

5

6

7

Value

0

1

0

0

0

0

1

0

Just flip the bit in position 0 to get the answer for the leftmost byte.

Bit

0

1

2

3

4

5

6

7

Value

1

1

0

0

0

0

1

0

This is X‘C2’.  The answer to this part is X‘C27B C000

Convert the following numbers to their representation in packed decimal.
      Give the hexadecimal representation with the proper number of hexadecimal digits.

      a)      123.75

      b)      –123.7

ANSWER:     a)   12375 has five digits.  It is represented as 12 37 5C.

                        b)   1237 has four digits.  Expand to 01237 and represent as 01 23 7D.

20  Give the correct Packed Decimal representation of the following numbers.

               a)   31.41         b)   –102.345            c)   1.02345

ANSWER:  Recall that the decimal is not stored, and that we need to have an odd
                     count of decimal digits.

      a)   This becomes                  3141,         or 03141.           03141C

      b)   This becomes              102345,      or 0102345       0102345D

      c)   This also becomes      102345,      or 0102345      0102345C.

21  Perform the following sums of numbers in Packed Decimal format. Convert to
               standard integer and show your math.  Use Packed Decimal for the answers.

      a)      025C + 085C             d)         666D + 444D

      b)      032C + 027D             e)         091D + 0C

      c)      10003C + 09999D

ANSWER:        Just do the math required and convert back to standard
                           Packed Decimal format.

      a)      025C + 085C represents 25 +85 = 110.  This is represented as 110C.

      b)      032C + 027D represents 32 –27 = 5.  This is represented as 5C.

      c)      10003C + 09999D represents 10003 –9999 = 4.  This is represented as 4C.

      d)      666D + 444D represents –666 –444 = –1110.  This is represented as 01 11 0D.

      e)      091D + 0C represents –91 +0 = –91  This is represented as 091D.

22  These questions concern 10–bit integers, which are not common.

      a)      What is the range of integers storable in 10–bit unsigned binary form?

      b)      What is the range of integers storable in 10–bit two’s–complement form?

      c)      Represent the positive number 366 in 10–bit two’s–complement binary form.

      d)      Represent the negative number –172 in 10–bit two’s–complement binary form.

      e)      Represent the number 0 in 10–bit two’s–complement binary form.

ANSWER:  Recall that an N–bit scheme can store 2N distinct representations.
                     For unsigned integers, this is the set of integers from 0 through 2N – 1.
                     For 2’s–complement, this is the set from – (2N–1) through 2N–1 – 1.

      a)      For 10–bit unsigned the range is 0 though 210 – 1, or 0 through 1023.

      b)      For 10–bit 2’s–complement, this is – (29) through 29 – 1, or – 512 through 511.


 

      c)                            366 / 2      = 183      remainder = 0
                                     183 / 2        = 91      remainder = 1
                                       91 / 2        = 45      remainder = 1
                                       45 / 2        = 22      remainder = 1
                                       22 / 2        = 11      remainder = 0
                                       11 / 2          = 5      remainder = 1
                                         5 / 2          = 2      remainder = 1
                                         2 / 2          = 1      remainder = 0
                                         1 / 2          = 0      remainder = 1.   
READ BOTTOM TO TOP!

The answer is 1 0110 1110, or 01 0110 1110, which equals 0x16E.
0x16E = 1·256 + 6·16 + 14 = 256 + 96 + 14 = 256 + 110 = 366.

The number is not negative, so we stop here.  Do not take the two’s complement unless the
number is negative.

      d)                            172 / 2        = 86      remainder = 0
                                       86 / 2        = 43      remainder = 0
                                       43 / 2        = 21      remainder = 1
                                       21 / 2        = 10      remainder = 1
                                       10 / 2          = 5      remainder = 0
                                         5 / 2          = 2      remainder = 1
                                         2 / 2          = 1      remainder = 0
                                         1 / 2          = 0      remainder = 1.   
READ BOTTOM TO TOP!

This number is 1010 1100, or 00 1010 1100, which equals 0x0AC.
0xAC = 10·16 + 12 = 160 + 12 = 172.

The absolute value:                       00 1010 1100
Take the one’s complement:        
11 0101 0011
Add one:                                      
          1
The answer is:                              
11 0101 0100     or 0x354.

      e)      The answer is                 00 0000 0000.
               You should just know this one.

23  These questions IBM Packed Decimal Form.

      a)      Represent the positive number 366 as a packed decimal with fewest digits.

      d)      Represent the negative number –172 as a packed decimal with fewest digits.

      e)      Represent the number 0 as a packed decimal with fewest digits.

ANSWER:     a)         366C
                        b)        
172D
                        c)        
0C       (not 0D, which is incorrect)