IBM Mainframe Forum

by **helen2000** » Sun Mar 14, 2010 11:38 pm

thanks Mr. Dick and Robert. by the way, what my understanding is: move num0 to num1, actually, exist
a assmeber statement does like that: nnnnbb or x'0000000000f0' ==>num1, so we got nnnnb0, is that right?
(I'm sorry, maybe I need learn assemble language right now :oops:

),
Mr. Robert? I don't know how to check the assembler statement in our output, do you have some idea, I can
try that if you have good idea. thanks very very much!

have a good day!

helen

by **dick scherrer** » Mon Mar 15, 2010 3:24 am

Hi Helen,

I'm sorry, maybe I need learn assemble language right now

At this point i suspect that being able to read some assembler would be helpful

These links are to a Princilpes of Operations (The Bible

), a site that shows many assembler instruction examples, and an Introduction for Assembler.
http://publibz.boulder.ibm.com/cgi-bin/ ... 0504121320

http://www.simotime.com/asmins01.htm

http://ulita.ms.mff.cuni.cz/pub/predn/s ... BM_390.pdf

actually, exist a assmeber statement does like that: nnnnbb or x'0000000000f0' ==>num1, so we got nnnnb0, is that right?

Not quite, i suspect. And depending on exactly which compiler with which options in effect, the generated code might be different.

I don't know how to check the assembler statement in our output

Specify LIST in the COBOL options for the compile. You will see something like:

Select all

000044 MOVE
00069A D205 8978 A0B1 MVC 2424(6,8),177(10) NUM0
000045 MOVE
0006A0 D201 897E C02A MVC 2430(2,8),42(12) NUM1
0006A6 D205 8980 8978 MVC 2432(6,8),2424(8) NUM1+2
0006AC 96F0 8985 OI 2437(8),X'F0' NUM1+7
000046 IF
0006B0 F247 D120 897E PACK 288(5,13),2430(8,8) TS2=0
0006B6 960F D124 OI 292(13),X'0F' TS2=4
0006BA F245 D128 8986 PACK 296(5,13),2438(6,8) TS2=8
0006C0 960F D12C OI 300(13),X'0F' TS2=12
0006C4 D504 D120 D128 CLC 288(5,13),296(13) TS2=0
0006CA 4770 B1B8 BC 7,440(0,11) GN=5(000
000047 DISPLAY
0006CE 5820 905C L 2,92(0,9) TGTFIXD+
0006D2 58F0 202C L 15,44(0,2) V(IGZCDS
0006D6 4110 A3B3 LA 1,947(0,10) PGMLIT A
0006DA 0DEF BASR 14,15

which is from the experiment using the 8-byte num1.

by **helen2000** » Tue Mar 16, 2010 3:03 am

I'm very appreciate Mr. Dick for detail answer and valuable assembler link. would you mind I ask for further question;

MOVE '9754' TO NUM0
MOVE NUM0 TO NUM1

I got relative assemble:

Select all

000191 MOVE
0008A2 D205 805A A0BB MVC 90(6,8),187(10) NUM0
000192 MOVE
0008A8 D205 8060 805A MVC 96(6,8),90(8) NUM1
0008AE 96F0 8065 OI 101(8),X'F0' NUM1+5

for the first assemble statement, MVC 90(6,8),187(10) , it's means: the 6 bytes content of offset (R10)+187 copy to offset (R8)+90;
could you tell me how to check R8 and R10 in my assemble code, I'm not sure if the following data area is helpful for the question. or
let me know if you need more information,thanks.

Select all

000148 (LIT+0) FFFFFFFC 00001000 00000001 00000000 5C5C40D4 D7E8D8F8 C6C3D
000168 (LIT+32) C1D9E340 D6C640E6 D6D9D2C9 D5C740E2 E3D6D9C1 C7C5405C 5CE3D
000188 (LIT+64) C5C3D6D9 C4E240D7 D9D6C3C5 E2E2C5E2 7AD5E4D4 F040C9E2 D57DE
0001A8 (LIT+96) C5D95AC3 D6D5E5C5 D9E2C9D6 D540C6C1 C9D35AD5 E4D4F040 C9E24
0001C8 (LIT+128) C9C3E2E4 C3C3C5E2 E2C6E4D3 5AE2E8E2 D6E4E340 40C9C7E9 E2D9E
0001E8 (LIT+160) F27E40F9 F7F5F440 40C8F2C4 C3C2D3D5 E4D4F17E D5E4D4F0 7EE6C
000208 (LIT+192) D51F1010 10101010 10101010 10101010 10101010 10101010 10101
000228 (LIT+224) 10101010 10101010 10101010 10101010 10101010 10101010 10101
000248 (LIT+256) 10101010 10101010 10101010 10101010 10101010 10101010 10101
000268 (LIT+288) 10101010 10101010 10101010 10101010 10101010 10101010 10101
000288 (LIT+320) 10101010 10101010 10101010 10101010 10101010 10101010 10101
0002A8 (LIT+352) 10101010 10101010 10101010 10101010 10101010 10101010 10101
0002C8 (LIT+384) 10108080 80808080 80808080 10101010 10108080 80808080 80808
0002E8 (LIT+416) 10101010 10101010 10101010 10101010 10100000 00000000 00000
000308 (LIT+448) 10100000 01340000 00020000 012C0000 00010000 01300000 00010

by the way, what does "LIT" means?
helen

by **dick scherrer** » Tue Mar 16, 2010 3:11 am

Hi Helen,

I'm not sure if the following data area is helpful for the question. or let me know if you need more information,thanks.
by the way, what does "LIT" means?

For this question, the data area (the program's literals) not so much. LIT in this case is for "literal" - as in Literal Pool:

Select all

LITERAL POOL MAP FOR LITERALS IN THE CGT:

000188 (LIT+0) FFFFFFFC 00001000 40000000 00000328 00000001 FFFFFFF
0001A8 (LIT+32) 00000262 00404400 00000248 00000229 00000068 0000021
0001C8 (LIT+64) 0000017E 00000159 00400000 000001C8 00000139 0040040

Notice the line just before the "data area" you posted.

I know of no way to check register content from a complie/assembly listing. To verify content of registers (and storage locations) a dump would be needed.

by **dick scherrer** » Tue Mar 16, 2010 3:21 am

Add on. . .

Keep in mind that register content is only valid for a very brief moment. Register content/values are modified very often.

What i believe you would be most interested in is the data contained at the addresses contained in certain registers.

by **Robert Sample** » Tue Mar 16, 2010 3:26 am

Two points to consider:
1. Even though you are moving '9754' to NUM0, notice that the assembler instruction generated is moving six bytes. The two spaces are in the literal pool at the end of the 9754. Alphanumeric moves, unless you are using reference modification, move the length of the data field whether or not your source is that long.

2. After moving the same six bytes to NUM1, the OI assembler instruction changes the last byte (offset NUM1+5) to a hex 'F0' which is a zero. COBOL is guaranteeing the value is unsigned by doing so.

by **helen2000** » Tue Mar 16, 2010 9:10 am

thanks so much, Mr. Dick and Robert,

What i believe you would be most interested in is the data contained at the addresses contained in certain registers

exactly correct! for the assmebler statement: MVC 90(6,8),187(10) NUM0
I want to check if the address 187(10) is 9754bb.

To verify content of registers (and storage locations) a dump would be needed

could you give me a sample about dump so that I can follow you to verify the register R10 and R8?
Wow, thanks for GOD, I got your help, otherwise I can't sleep tonight.

Helen

by **dick scherrer** » Tue Mar 16, 2010 9:21 am

Hi Helen,

One way to generate a dump is to force an abend (i.e. cause a s0c7)

Another is to generate an LE (Language Environment) dump - look here:
http://publibz.boulder.ibm.com/cgi-bin/ ... g50/1.13.1

by **helen2000** » Wed Mar 17, 2010 3:04 am

I'm sorry, Mr. Dick. I tried lots of ways but I can't get the S0c7. I always get compile failure once I modify any code,
could you give me some detail, thanks,
have a good day!
helen

by **Robert Sample** » Wed Mar 17, 2010 4:31 am

I thought usually the problem was programmers could not get rid of S0C7 abends ... sigh ... this code will do generate one:

Select all

WORKING-STORAGE SECTION.
01 WS-DATA.
05 WS-VAR1 PIC X(04) VALUE SPACES.
05 WS-VAR2 REDEFINES WS-VAR1
PIC S9(04).
*
PROCEDURE DIVISION.
S1000-INITIALIZE.
ADD 1 TO WS-VAR2.
GOBACK.

IBM Mainframe Forum

how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?

Re: how to convert string to number?