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 hexadecimal to decimal, are often best done via binary.

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.

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. Note that the hexadecimal numbering system calls for more than ten digits,
so we use the first six letters of the alphabet.

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}

The fact that the bases for octal and hexadecimal are powers of the basis for binary facilitates the
conversion between these bases. 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 octal and hexadecimal is best done by first converting to binary.

Except for an occasional reference, we shall not use the octal system much, but focus on the decimal,
binary, and hexadecimal numbering systems.

The figure below shows the numeric equivalents in binary, octal, and decimal of the first 16 hexadecimal
numbers. If octal numbers were included, they would run from 00 through 017.

Binary Decimal Hexadecimal
(base 2) (base 10) (base 16)

0000 00 0

0001 01 1 Note that conversions from hexadecimal

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

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

0100 04 4 E = 1110. We shall normally denote

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

0110 06 6 to facilitate reading the binary.

0111 07 7

1000 08 8 Conversion from binary to hexadecimal

1001 09 9 is also quite easy. Group the bits four at

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

1011 11 B Thus 10111101, written 1011 1101 for

1100 12 C clarity is BD because 1011 = B and

1101 13 D 1101 = D.

1110 14 E

1111 15 F

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, thus obtaining 0011 1010,
which is converted to 3A as 0011 = 3 and 1010 = A.

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 an arbitrary
member of that infinite set. Computer storage formats allow only for the representation of a large,
but finite, subset of the integers.

It is easy to show that an N–bit binary integer can represent one of 2^N 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 B_NB_N–1 … B₃B₂B₁.
By the inductive hypothesis, this can represent one of 2^N possible values.

3. We now consider an (N+1)–bit integer, written as B_N+1B_NB_N–1 … B₃B₂B₁.
By the inductive hypothesis, there are 2^N values of the form 0B_NB_N–1 … B₃B₂B₁,
and 2^N values of the form 1B_NB_N–1 … B₃B₂B₁.

4. The total number of (N+1)–bit values is 2^N + 2^N = 2^N+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
2^N integer values in the range 0 through 2^N – 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.

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 in binary.
As before we present two proofs and let the student choose his/her favorite and ignore the other.

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 top to
bottom as least significant bit to most significant bit. 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. 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

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.

We now develop a “power of 2” notation that will be required when we study the IEEE floating
point standard. We have just shown that decimal 23.375 = 10111.011 binary. Recall that in the
scientific “power of 10” notation, when we move the decimal to the left one place we have to
multiply by 10. Thus, 1234 = 123.4 · 10¹ = 12.34 · 10² = 1.234 · 10³.

We apply the same logic to the binary number. In the IEEE standard we need to form the number
as a normalized number, which is of the form 1.xxx · 2^p. In changing 10111 to
1.0111 we have moved the decimal point (O.K. – it should be called binary point) 4 places to
the left, so 10111.011 = 1.0111011 · 2⁴. Recalling that 2⁴ = 16 and 2⁵ = 32, and noting that
16.0 < 23.375 £ 32.0 we see that the result is as expected.

Conversion from binary to decimal is quite easy. One just remembers the decimal representations
of the powers of 2. We convert 10111.011 binary to decimal. Recalling the positional notation
used in all number systems:

10111.011 = 1·2⁴ + 0·2³ + 1·2² + 1·2¹ + 1·2⁰ + 0·2^-1 + 1·2^-2 + 1·2^-3

= 1·16 + 0·8 + 1·4 + 1·2 + 1·1 + 0·0.5 + 1·0.25 + 1·0.125
= 23.375

Conversions between Decimal and Hexadecimal

The conversion is best done 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

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.

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 / 10^K 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
10^K = 2^K·5^K. 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 = 2^K·5^K. This has an integer solution J only
if the right hand side of the equation can be factored by 3. But neither 2^K nor 5^K can be
factored by 3, so the right hand side cannot be factored by 3 and hence the equation is not solvable.

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 = 2^K. 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 numbers.

Signed and Unsigned Integers

Fixed point numbers include real numbers with a fixed number of decimals, such as those
commonly used to denote money amounts in the United States. We shall focus only on
integers and relegate the study of real numbers to the floating point discussion.

Integers are stored in a number of formats. The most common formats today include 16
and 32 bits. The new edition of Visual Basic will include a 64–bit 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
	Sign	MSB		LSB

Bit #	0	1	2 – 6	7
	Sign	MSB		LSB

The simplest topic to cover is the storage of unsigned integers. As there are 2^N possible
combinations of N binary bits, there are 2^N unsigned integers ranging from 0 to 2^N – 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; in one’s complement, decimal –100 = 1001 1011 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.

The Two’s Complement
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, and we
avoid the messy problem of having – 0 ¹ 0.

In N- bit two’s complement arithmetic, + 127 0111 1111
the range of integers that can be + 10 0000 1010
represented is – 2^N-1 through 2^N-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 high
binary representations for this example. – 128 1000 0000 order 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 = (10^0.30103)⁶³ = 10^18.9649 = 10^0.9649 · 10¹⁸ = 9.224 · 10¹⁸, so a 64-bit integer allows
the representation of 18 digit numbers and most 19 digit numbers. The precise range for
64–bit signed integers 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 R_k = A_k 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, which evaluates to 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, evaluating as 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 ... 2^N – 1, inclusive.
For N-bit two’s-complement integers the range of non-negative integers representable is

0 ... 2^N-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 shall work this by relying on the fact that,
if B = –A, then A + B = 0.

We begin with a few issues of notation. Beginning with the next chapter of the textbook,
the plus sign, “+”, will be used mostly to denote the logical OR operation. For the purposes
of this discussion, it will denote addition. In this discussion, the symbol will denote the
one’s–complement of the single–bit number a. In later chapters, this will be used to refer
to the logical NOT of the Boolean value a. In fact, the two uses are identical; one generates
the one’s–complement of a single bit by passing it through a NOT gate.

We now consider the bitwise addition of a binary number and its one’s–complement. At the
bit level, the addition table is simple, and can be represented by a half–adder.

A Sum Carry
0 1 1 0
1 0 1 0

Note that there are only two rows in this table, because the value of a binary bit determines
the value of its one’s–complement.

We next prove a mathematical theorem to be used in our demonstration that the
two’s–complement of an integer is, in fact, its negative.

Theorem: For N ≥ 0, we have

Proof: We display this by simple mathematical induction.
We give two base cases: 1 = 2⁰ = 2¹ – 1, and 1 + 2 = 2⁰ + 2¹ = 3 = 2² – 1.

Now if 1 + 2¹ + … + 2^N = 2^N+1 – 1, then
1 + 2¹ + … + 2^N + 2^N+1 = (2^N+1 – 1) + 2^N+1 = 2·2^N+1 – 1 = 2^N+2 – 1.

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.

Let A be the N–bit binary number represented as A_N–1A_N–2 … A₂A₁A₀.

The unsigned decimal value represented by this string is given by

.

For example, consider the 8–bit unsigned integer A = 0110 0100.
A₇ = 0, A₆ = 1, A₅ = 1, A₄ = 0, A₃ = 0, A₂ = 1, A₁ = 0, and A₀ = 0;
the value is 0·2⁷ + 1·2⁶ + 1·2⁵ + 0·2⁴ + 0·2³ + 1·2² + 0·2¹ + 0·2⁰ =
2⁶ + 2⁵ + 2² = 64 + 32 + 4 = 100.

We now consider N–bit patterns, interpreted as two’s–complement integers. In this form,
the bit pattern A_N–1A_N–2 … A₂A₁A₀ represents the value given by the sum

Again, consider the 8–bit unsigned integer A = 0110 0100.
A₇ = 0, A₆ = 1, A₅ = 1, A₄ = 0, A₃ = 0, A₂ = 1, A₁ = 0, and A₀ = 0;
the value is – 0·2⁷ + 1·2⁶ + 1·2⁵ + 0·2⁴ + 0·2³ + 1·2² + 0·2¹ + 0·2⁰ =
2⁶ + 2⁵ + 2² = 64 + 32 + 4 = 100.

For example, consider the 8–bit unsigned integer A = 1001 1100.
A₇ = 1, A₆ = 0, A₅ = 0, A₄ = 1, A₃ = 1, A₂ = 1, A₁ = 0, and A₀ = 0;
the value is –1·2⁷ + 0·2⁶ + 0·2⁵ + 1·2⁴ + 1·2³ + 1·2² + 0·2¹ + 0·2⁰ =
–1·2⁷ + 1·2⁴ + 1·2³ + 1·2² = –128 + 16 + 8 + 4 = –100.

Here is a proof that the two’s–complement represents the negative of a number. We
shall have a bit more to say about the proof after we give it. Let A be represented by:

If A_N–1=1, the number is negative. If A_N–1=0, the number is not negative.

Let B = – A, as represented in the two’s–complement notation. B represents the value:

But we have shown that for every bit index that

So we have the following.

In order to illustrate this more fully, we consider the negative range for four–bit
integers represented in two’s–complement form. The range is from –8 to –1 inclusive.

Here is the table with the results for discussion.

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.

For the adder, we have just the following, assuming that B = – A.

But the 2^N represents a carry–out from the high–order (or sign) bit; it is not part of the
sum, which is still 0. In order to see this, consider the sum 100 + (–100), considered
as the sum of two eight–bit two’s–complement integers.

+100 = 0110 0100
-100 = 1001 1100
Sum is 1 0000 0000

The eight–bit sum is still 0.

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 · 2^P, 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 · 2^P, 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 · 2^P, 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: Avagadro’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 Avagadro’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.

Avagadro’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
2^X = (10^0.3010300)^X = 10.^23.7797520 or 0.30103·X = 23.7797510 for X = 78.9946218.

If we use N_A to denote Avagadro’s number, the first thing we have discovered from this tedious
analysis is that 2⁷⁸ < N_A < 2⁷⁹, and that N_A » 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 = (10^0.30103)⁷⁸ = 10^23.48034 = 10^0.48034 · 10²³ = 3.022317 · 10²³.
But 6.022142 / 3.022317 = 1.992558, so N_A = 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. Loosely speaking, we say 1 / 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.

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 CPSC 2105 mid-term exam and
the paragraphs following were taken from the answer key for the 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.

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 Avagadro’s number.
We have seen that N_A = 1.1111 1110 0001 · 2⁷⁸. This is a positive number, so 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 10^0.53 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.

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.

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 (10^0.30103)²⁵² = 10^75.86 » 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 2^F.

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 » (10^0.30103)⁵⁶ = 10^16.85 » 9·10¹⁶.
2¹²⁰ » (10^0.30103)¹²⁰ = 10^36.12 » 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 < 9·10^75	7 digits
Double Precision	3.0·10^–77 < X < 9·10^75	16 digits
Extended Precision	3.0·10^–77 < X < 9·10^75	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·16^E, 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·16^E, 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·16^E, 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 shall 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.

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.

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 – 3	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
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	Letterlike 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

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 2^N distinct representations.
For unsigned integers, this is the set of integers from 0 through 2^N – 1.
For 2’s–complement, this is the set from – (2^N–1) through 2^N–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)