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Amit Manas Dubey
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Hi
Can somebody tell me how can we calculate the space utilized by COMP variables in COBOL ?
For example S9(8) COMP - Takes 4 bytes ...Is there a way this can be calculated ?? Also, is there a diff in the no of bytes used by signed and unsigned numeric variables.
Thanks
Amit |
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William Thompson
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| Amit Manas Dubey wrote: |
| Can somebody tell me how can we calculate the space utilized by COMP variables in COBOL ? For example S9(8) COMP - Takes 4 bytes ...Is there a way this can be calculated ?? |
Comp is binary and the assembler opcode that does binary arithmetic are register instructions. Register instructions either deal with full words or half words. The registers themselves are full words. Full words are 4 bytes long.
| Quote: |
| Also, is there a diff in the no of bytes used by signed and unsigned numeric variables. |
In binary (comp) fields, only the one high bit signifies the sign, no extra length needed.
In packed (comp3) fields the last half byte signifies the sign, always there and no extra length needed. |
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Amit Manas Dubey
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I saw the manuals....As I understand say S9(4) COMP - can have 9999 as Max value - this when converted to Binary is 10011100001111 + 1 bit for sign = 15 bits .....This will take 2 bytes = 16 bits........
Is my understanding correct ? |
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William Thompson
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| Amit Manas Dubey wrote: |
I saw the manuals....As I understand say S9(4) COMP - can have 9999 as Max value - this when converted to Binary is 10011100001111 + 1 bit for sign = 15 bits .....This will take 2 bytes = 16 bits........
Is my understanding correct ? |
Kinda.....Some compiler (or LE) thingy will force the max number to be no greater than the number of digits specified - S9(4) means 9999 is max. But without that setting, 32767 (7FFF) is possible. |
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Amit Manas Dubey
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Pls explain a little more...You mean to say that the storage can be coded in Hexadecimal as well...Okies....why not FFFF ?? is it because of signed stuff - 32767 - 0 - + 32767 ?
Pls elaborate !! |
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William Thompson
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| Amit Manas Dubey wrote: |
Pls explain a little more...You mean to say that the storage can be coded in Hexadecimal as well...Okies....why not FFFF ?? is it because of signed stuff - 32767 - 0 - + 32767 ?
Pls elaborate !! |
Actually it is -32768 (FFFF) +32767 (7FFF) - remember the high bit is the sign.
comp-val pic s9(4) comp.
hex-val redefines comp-val pic x(4).
move x'1000' to hex-val.
display comp-val.
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Amit Manas Dubey
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Thanks a lot Tom...
I am clear on this now...Just curious to know - how do you revert so fast on these posts ? unless u are addicted to this now !!  |
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William Thompson
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I'm not quite that fast....  |
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sandip_mainframe
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hi,
usage comp it is used for numeric value stored in pure binary form.
PIC S9(1) TO PIC S9(4) IT TAKES 2 BYTES = HALF WORD = 16 BITS
PIC S(5) TO PIC S9(9) IT TAKES 4 BYTES = FULL WORD = 32 BITS
PIC S9(10) TO PIC S9(19) IT TAKES 8 BYTES =DOUBLE WORD = 64 BITS.
SANDIP WALSINGE. |
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Selvamani
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Sandeep,
I remember that 1 byte = 4 bits. If this is correct, Then your calculation needs to be relooked.
Am I correct? |
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William Thompson
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| Selvamani wrote: |
| I remember that 1 byte = 4 bits. |
No, one byte is 8 bits
no  |
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dick scherrer
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Hello,
1 byte = 8 bits. . .
For example the letter A is made up of these bits
1100 0001 which is read as C1. The numeral 9 is 1111 0101 (F9). |
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UmeySan
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Location: Germany
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Hi !
Coming back to main question. As Sandip wrote it's like:
PIC 9 to 9(4) -> 2 Bytes
PIC 9(5) to 9(9) -> 4 Bytes
PIC 9(10) to 9(18) -> 8 Bytes
With binary, you only have halfbyte, fullbyte and douplebyte.
@Sandip: S9(19) ??? Are you really shure ???
Just a little tip. Code alle your desired field in al little programm.
Like PIC 9(01) comp / PIC 9(02) comp / and so on.
Compile your programm. Then have a look at the listing.
PIC S9(9) COMP. BLW=00000+F45,0000035 4C
So here you could see, that S9(9) resides in a four byte field. Just 4C.
Regards, Umeysan |
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