IBM Mainframe Forum

by **utpalpal07** » Sat Jul 21, 2012 10:34 pm

Hi Guys,

   PACK W_A,BYTES
   MP W_A,=P'1024'
   UNPK W_B,W_A
W_BYTES DC P'373'
W_A    DC P'0000000'
W_B    DC P'0000000'

I"m multiplying the 2 variables.
But i'm getting o/p as: 38195B, where as i'm expecting: 381952

Its considering last digit as +2 so its converting as 'B" as signed clause.

How to get a o/p without signed clause?

Please help

Regards
Utpal

by **steve-myers** » Sat Jul 21, 2012 11:22 pm

There is no such animal as an unsigned packed decimal number. Period. End of story. The 4 byte, 8 packed decimal digits time of day returned by the TIME macro is not a true packed decimal number, and you cannot used packed decimal arithmetic operations on it.
You did not bother to show us what is in BYTES. How can we evaluate any other claims you have made?

by **steve-myers** » Sat Jul 21, 2012 11:40 pm

3. Read, carefully, the definition of the PACK and UNPK instructions in Principles of Operations. The last byte in the input is treated differently than the other bytes. The output you show is perfectly valid when the input is a packed decimal 381952. There are two methods you can use to correct this. Please tell us at least one of them when you figure it out. If you are correct I will tell you the other one.

by **BillyBoyo** » Sun Jul 22, 2012 4:54 am

But i'm getting o/p as: 38195B, where as i'm expecting: 381952

Where are you "getting" this? The format of output you have shown is Signed Zoned Decimal. You have no leading zero, so you must also have "edited" to drop that. A clue.

Really read the Principles of Operation. Concentrate also on the terminology, I've never come across a "signed clause".

It is also important to realise that "it" didn't do this, or do anything. Computers do exactly what we tell them with our code. We make mistakes. "It" just applies our mistakes to data. If you continue to wonder about "it" being wrong, you will waste a lot of time.

by **steve-myers** » Sun Jul 22, 2012 7:54 am

Actually, utpalpal07 lied to us. His W_B is too small, as shown, to hold the output he claims he got.

by **steve-myers** » Sun Jul 22, 2012 8:26 am

After a little more checking, utpalpal07 never ran the alleged code. I prepared and ran this -

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000000 00000 00023 1 UTPAL CSECT
R:F 00000 2 USING *,15
000000 FC32 F012 F020 00012 00020 3 MP W_A,=P'1024'
000006 F353 F016 F012 00016 00012 4 UNPK W_B,W_A
00000C 0700 5 FREEZE NOPR 0
00000E 1BFF 6 SR 15,15
000010 07FE 7 BR 14
000012 0000373C 8 W_A DC P'0000373'
000016 F0F0F0F0F0F0 9 W_B DC C'000000'
000020 10 LTORG ,
000020 01024C 11 =P'1024'
000000 12 END UTPAL

It got an S0C7 executing the MP instruction. It got the 0C7 because W_A is not large enough.

by **steve-myers** » Tue Jul 24, 2012 1:58 am

utpapa07: if you send us something to analyze, it should be as correct as possible except for the piece you want us to analyze. For example:

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UTPAL CSECT
USING *,15
L 0,=A(373*1024)
CVD 0,W_A
UNPK W_B,W_A
SR 15,15
BR 14
DC 0D'0'
W_A DC PL8'0'
LTORG ,
W_B DC C'000000'
END UTPAL

demonstrates the issue you want discussed, though some might complain W_B is too short for W_A, it does not affect the execution of the UNPK instruction, which is the real issue.

Now only you know how you really prepared W_A, but we all know the code you showed us was not how it was prepared.

Don't try to fool us. It won't work, and it delays what you really want to know.

by **utpalpal07** » Mon Jul 30, 2012 7:34 pm

Hello Steve,

I din lied but found the solution also.
here's the code:

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ADD CSECT
STM 14,12,12(13) SAVE THE REGISTERS
LR 12,15 MEKE R12 THE BASE REGISTER
USING ADD,12 FINISH THE JOB
LA 11,REGSAVE POINT TO SAVE AREA
ST 13,4(11) SAVE CURRENT R13
ST 11,8(13) INFORM THE OS
LR 13,11 ESTABLISH NEW R13
*
WTO 'MULTIPLICATION'
MP BYTES,=P'1024'
UNPK W_A,BYTES
OI W_A+6,X'F0'
MVC WTOA+8(9),W_A PRINT OUTLINE
WTOA WTO 'XXXXXXXXX IS W_A'
*
*
L 13,4(13) RESTORE R13 TO ORIGINAL
LM 14,12,12(13) RESTORE THE REGS
SR 15,15 EVERYTHING WENT OK
BCR B'1111',14 RETURN TO THE SYSTEM
*
BYTES DC PL5'373'
W_A DC PL7'0000000'
W_B DC PL7'0000000'
REGSAVE DS 18F
END

Compiled steps:

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000000 00000 000CB 1 ADD CSECT
000000 90EC D00C 0000C 2 STM 14,12,12(13)
000004 18CF 3 LR 12,15
R:C 00000 4 USING ADD,12
000006 41B0 C080 00080 5 LA 11,REGSAVE
00000A 50DB 0004 00004 6 ST 13,4(11)
00000E 50BD 0008 00008 7 ST 11,8(13)
000012 18DB 8 LR 13,11
9 *
10 WTO 'MULTIPLICATION'
000014 12+ CNOP 0,4
000014 A715 000B 0002A 13+ BRAS 1,IHB0001A
000018 0012 14+ DC AL2(18)
00001A 0000 15+ DC B'0000000000000000'
00001C D4E4D3E3C9D7D3C9 16+ DC C'MULTIPLICATION'
00002A 17+IHB0001A DS 0H
00002A 0A23 18+ SVC 35
00002C FC42 C06A C0C8 0006A 000C8 19 MP BYTES,=P'1024'
000032 F364 C06F C06A 0006F 0006A 20 UNPK W_A,BYTES
000038 96F0 C075 00075 21 OI W_A+6,X'F0'
00003C D208 C04C C06F 0004C 0006F 22 MVC WTOA+8(9),W_A
23 WTOA WTO 'XXXXXXXXX IS W_A'
000042 0700 25+ CNOP 0,4
000044 A715 000C 0005C 26+WTOA BRAS 1,IHB0003A
000048 0014 27+ DC AL2(20)
00004A 0000 28+ DC B'0000000000000000'
00004C E7E7E7E7E7E7E7E7 29+ DC C'XXXXXXXXX IS W_A'
00005C 30+IHB0003A DS 0H
00005C 0A23 31+ SVC 35
32 *
00005E 58DD 0004 00004 33 L 13,4(13)
000062 98EC D00C 0000C 34 LM 14,12,12(13)
000066 1BFF 35 SR 15,15
000068 07FE 36 BCR B'1111',14
37 *
00006A 000000373C 38 BYTES DC PL5'373'
00006F 0000000000000C 39 W_A DC PL7'0000000'
000076 0000000000000C 40 W_B DC PL7'0000000'
000080 41 REGSAVE DS 18F
42 END
0000C8 01024C 43 =P'1024'

Output:

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MULTIPLICATION
0381952 IS W_A

Please have a look. And i Just asked how to remove the signed bit in the last.

Regards
Utpal

by **steve-myers** » Mon Jul 30, 2012 9:18 pm

Well, you still keep posting code that does not match your alleged execution. This time you moved a 9 byte field in your WTO macro from a 7 byte field, but your alleged output just shows 7 bytes.

In any event,

OI W_A+6,X'F0'

is correct, but something like this is better.

OI W_A+L'W_A-1,X'F0'

The other way is to adjust the packed decimal sign before the UNPK.

OI BYTES+L'BYTES-1,X'0F'
UNPK W_A,BYTES

A little more complicated in some respects, but you don't have to fiddle with fixing packed decimal signs, is to use the ED instruction. The down side of ED is you must exactly (something you appear reluctant to do) match the output mask with the digits in the input. This means your 5 byte (9 digit) BYTES must be edited to an output mask that has 9 digit select characters.

ED OUTPUT,BYTES
...
OUTPUT DC X'40202020202020202120'
BYTES DC P'123456789'

The first byte in OUTPUT is a fill character that is inserted in place of leading 0 characters. The 40 character is an EBCDIC blank. The 20 and 21 characters are "digit select" characters that are replaced with the fill character or a digit from the packed decimal number. The 21 character is a "significance start"that tells ED that leading 0s are inserted into the output after the character. Using the edit mask I showed, PL5'0' will produce

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----+----1
C' 0'

by **utpalpal07** » Mon Jul 30, 2012 9:58 pm

Thanks steve for the help and description. Its useful

IBM Mainframe Forum

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