Printing
Packed Data
The
unpack command, UNPK, has an unfortunate side effect.
Two
commands to convert binary to printable EBCDIC.
CVD Converts
the binary to packed decimal.
UNPK Almost
converts to EBCDIC.
Consider
the decimal number 42, represented in binary in register R4.
CVD
R4,PACKOUT produces the packed decimal 042C.
When
this is unpacked it should become F0
F4 F2
Unpack
just swaps the sign half byte: F0
F4 C2.
This
prints as 04B, because 0xC2 is the EBCDIC code for the letter ‘B’.
We
have to correct the zone part of the last byte.
See pages 167 & 175.
Printing
Packed Data (Part 2)
Here
is the code that works.
NUMOUT CVD R4,PACKOUT CONVERT THE NUMBER TO PACKED
UNPK THESUM,PACKOUT
PRODUCE FORMATTED NUMBER
MVZ THESUM+7(1),=X’F0’ MOVE 1 BYTE
* TO ADDRESS
THESUM+7
BR 8 RETURN ADDRESS IN REGISTER 8
PACKOUT DS PL8 HOLDS THE PACKED OUTPUT
THESUM has eight
characters stored as eight bytes. The
addresses are:
SUM 
SUM +1 
SUM +2 
SUM +3 
SUM +4 
SUM +5 
SUM +6 
SUM +7 





Hundreds 
Tens 
Units 
Again, the expression THESUM+7 is an address,
not a value.
If THESUM holds C‘01234567’,
then THESUM+7 holds C ‘7’.
A Problem
with the Above Routine
Consider
the decimal number –42, stored in a register
in binary two’s–complement form.
CVD produces 042D
UNPK produces F0
F4 D2
The
above MVZ will convert this to F0 F4 F2, a positive number.
There
are some easy fixes that are guaranteed to produce the correct
representation for a negative number.
Most
of the fixes using CVD and UNPK depend on placing the minus sign
to the right of the digits. So that the negative integer –1234 would be
printed as “1234–”.
My Version
of NUMOUT (Number Out)
This
routine avoids packed decimal numbers.
We
are given a binary number (negative or non–negative) in register R4.
1. Is the number negative?
If so, set the sign to ‘–’ and take
the absolute value.
Otherwise, leave the sign as either
‘+’ or ‘ ’.
We now have a non–negative number. Assume it is not zero.
2. Divide the number by 10,
get a quotient and a remainder.
The remainder will become the
character output.
3. The remainder is a positive number in the range
[0, 9].
Add =X‘F0’ to produce the EBCDIC
code.
4. Place this digit code in the proper output
slot.
Is the quotient equal to 0? If so, quit.
If it is not zero, place the quotient in the dividend
and return to 2.
NUMOUT:
Example
Consider the positive integer 9413. We want to print this out.
Do repeated division by 10 and watch the remainders.
9413 divided by 10: Quotient
= 941 Remainder = 3. Generate digit “3”.
941 divided by 10: Quotient
= 94 Remainder = 1. Generate digit “1”.
94 divided by 10: Quotient
= 9 Remainder = 4. Generate digit “4”.
9 divided by 10: Quotient
= 0 Remainder = 9. Generate digit “9”.
Quotient is zero, so the process stops.
As they are generated, the digits are placed right to
left, so that the result
will print as the string 9413.
NUMOUT:
Specifications
The code processes a 32–bit two’s–complement integer,
stored as a fullword
in register R5 and prints it out as a sequence of EBCDIC characters.
The specification calls for printing out at most 10
digits, each as an EBCDIC
character. The sign will be placed in
the normal spot, just before the number.
For no particular reason, positive numbers will be
prefixed with a “+”.
I just thought I would do something different.
This will use repeated division, using the even–odd
register pair (R4, R5),
which contains a 64–bit dividend.
As a part of our processing we shall insure that the
dividend is a 32–bit
positive number. In that case, the “high
order” 32 bits of the number are all 0.
For that reason, we initialize the “high order”
register, R4, to 0 and initialize the
“low order” register, R5, to the absolute value of the integer to be output.
The EBCDIC characters output will be placed in a
12–byte area associated with
the label CHARSOUT, at byte addresses CHARSOUT through CHARSOUT+11.
Two New
Instructions: LCR and STC
The code below will use
two instructions that have not yet been discussed.
LCR (Load Complement Register)
Example LCR R1,R2
This loads
register R1 with the negative (two’s–complement) of the
value in register R2. This is also used
in my routine NUMIN.
STC (Store Character)
Example STC
R8,CHARSOUT(R3)
PLACE THE DIGIT
This transfers the EBCDIC
character, with code in the low order 8 bits of the
source register, to the target address.
None of the bits in the register are changed.
The idea behind NUMOUT is to
compute the numerical value of a digit in a
source register, convert it to an EBCDIC code, and move it to the print line.
NUMOUT: Part
1
The first part checks the sign of the integer in
register R4 and sets the
sign character appropriately.
NUMOUT MVC CHARSOUT,ZEROOUT DEFAULT TO 0
MVI THESIGN,C‘+’ DEFAULT TO A PLUS SIGN
C
R5,=F‘0’
COMPARE R5 TO 0
BE DONE VALUE IS 0, NOTHING TO DO
BH ISPOS VALUE IS POSITIVE
MVI THESIGN,C‘’ PLACE A MINUS SIGN
LCR R5,R5 2’S COMPLEMENT R5 TO MAKE POS
ISPOS SR R4,R4 CLEAR REGISTER 4
Here are
some data declarations used with this part of the code.
* 123456789012
ZEROOUT DC C‘ 0’
11 SPACES AND A ZERO
CHARSOUT DS CL12 UP TO 11 DIGITS AND A SIGN
Division
(Specifically D – Divide Fullword)
This instruction divides a 64–bit dividend, stored in
an even–odd register pair,
by a fullword, and places the quotient and remainder back into the register
pair.
This will use the even–odd register pair (R4,
R5). The specifics of the
divide instruction are as follows.

R4 
R5 
Before
division 
Dividend
(high order 32 bits) 
Dividend
(low order 32 bits) 
After
division 
Remainder 
Quotient 
There
are specific methods to handle dividends that might be negative.
As we are
considering only positive dividends, we ignore these general methods.
Our Example
of Division
Start with a binary number in register R5.
We assume that register R4 has been cleared to 0, as
this example
is limited to a 32–bit positive integer.
This code will later be modified to process the
remainder, and store
the result as a printable EBCDIC character.
DIVIDE D R4,=F‘10’ DIVIDE (R4,R5) BY TEN
*
* THE REMAINDER, IN
R4, MUST BE PROCESSED AND STORED
*
SR R4,R4 CLEAR R4 FOR
ANOTHER
C R5,=F‘0’ CHECK THE QUOTIENT
BH DIVIDE CONTINUE IF QUOTIENT > 0
Placing the
Digits
At
this point, our register and storage usage is as follows:
Register
R3 will be used as an index register.
Register
pair (R4, R5) is being used for the division.
Register
pair (R6, R7) is reserved for use by the BXH instruction.
CHARSOUT DS CL12 contains the twelve characters that form
the
print representation of the integer.
The
strategy calls for first placing a digit in the units slot (overwriting the ‘0’)
and then moving left to place other digits.
To allow for a sign, no digit
is to be placed in slot 0, at address CHARSOUT.
The
idea will be to place the character into a byte specified by CHARSOUT(R3).
The
register is initialized at 11 and decremented by 1 using the BXH instruction.
The Digit
Placement Code
Here is a sketch of the digit
placement code. It must be integrated
into
the larger DIVIDE loop in order to make sense.
The register pair (R6, R7) is
used for the BXH instruction.
R6 holds the increment value
R7 holds the limit value
L R6,=F‘1’ SET
INCREMENET TO 1
SR R7,R7 CLEAR R7.
LIMIT VALUE IS 0.
L R3,=F‘11’ SET
A
R4,=X‘F0’ ADD TO GET EBCDIC CODE FOR DIGIT
STC R4,CHARSOUT(R3) PLACE THE CHARACTER
BXH R3,R6,DIVIDE GO BACK TO TOP OF
MVC CHARSOUT(R3),
The Complete
Divide Loop
Here is the complete code for
the divide loop. Note the branch out of
the loop.
L R6,=F‘1’ SET INCREMENET TO 1
SR R7,R7 CLEAR R7. LIMIT VALUE IS 0.
L R3,=F‘11’ SET
*
DIVIDE D R4,=F‘10’ DIVIDE (R4,R5) BY TEN
A R4,=X‘F0’ ADD TO GET EBCDIC CODE FOR DIGIT
STC R4,CHARSOUT(R3) PLACE THE CHARACTER
SR R4,R4 CLEAR R4 FOR ANOTHER
C R5,=F‘0’ CHECK THE QUOTIENT
BNH PUTSIGN EXIT LOOP IF QUOTIENT <= 0
BXH R3,R6,DIVIDE GO BACK TO TOP OF
*
PUTSIGN MVC CHARSOUT(R3),