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COMP, COMP1, COMP2 & COMP3 in COBOL

 
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raghu_4412
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PostPosted: Sat Mar 06, 2004 8:17 am    Post subject: COMP, COMP1, COMP2 & COMP3 in COBOL
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hello sir,

please explain me about COMP, COMP1, COMP2 & COMP3 with an example? how is memory allocated in each of the above?

what is meant by halfword & fullword binary in COMP & how the memory allocates?


please clear me doubt about the above mentiones questions.
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sanga_1978

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PostPosted: Sun Mar 07, 2004 8:35 pm    Post subject: details about comp
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hai,


the following is the details of COMP items.


BINARY
Specified for binary data items. Such items have a decimal equivalent
consisting of the decimal digits 0 through 9, plus a sign. Negative
numbers are represented as the two's complement of the positive number
with the same absolute value.

The amount of storage occupied by a binary item depends on the number
of decimal digits defined in its PICTURE clause:

Digits in PICTURE Clause Storage Occupied
1 through 4 2 bytes (halfword)
5 through 9 4 bytes (fullword)
10 through 18 8 bytes (doubleword) │
The leftmost bit of the storage area is the operational sign.

PACKED-DECIMAL
Specified for internal decimal items. Such an item appears in storage
in packed decimal format. There are 2 digits for each character
position, except for the trailing character position, which is
occupied by the low-order digit and the sign. Such an item can
contain any of the digits 0 through 9, plus a sign, representing a
value not exceeding 18 decimal digits.

The sign representation uses the same bit configuration as the 4-bit
sign representation in zoned decimal fields (see Table 12 in
topic 2.7.12.2 and Table 13 in topic 2.7.12.2).

COMPUTATIONAL or COMP
Representation of the COMPUTATIONAL phrase is system-dependent and is
normally assigned to representations that yield the greatest
efficiency when arithmetic operations are performed on that system.
For the VS COBOL II compiler, the COMPUTATIONAL phrase is synonymous
with BINARY.

COMPUTATIONAL-1 or COMP-1
Specified for internal floating-point items (single precision).
COMP-1 items are 4 bytes long. The sign is contained in the first bit
of the leftmost byte and the exponent is contained in the remaining 7
bits. The last 3 bytes contain the mantissa.

COMPUTATIONAL-2 or COMP-2
Specified for internal floating-point items (double precision).
COMP-2 items are 8 bytes long. The sign is contained in the first bit
of the leftmost byte and the remaining 7 bits contain the exponent.
The remaining 7 bytes contain the mantissa.

COMPUTATIONAL-3 or COMP-3 (internal decimal)
For VS COBOL II, this is the equivalent of PACKED-DECIMAL.

COMPUTATIONAL-4 or COMP-4 (binary)
For VS COBOL II this is the equivalent of BINARY.

by
g.s.sanker
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thalapradheep

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PostPosted: Mon Jan 28, 2008 5:01 pm    Post subject: need help?
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How much bytes will S9(18) Comp will occupy?
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sri_mf

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PostPosted: Mon Jan 28, 2008 5:08 pm    Post subject: Reply to: COMP, COMP1, COMP2 & COMP3 in COBOL
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How much bytes will S9(18) Comp will occupy?


8 bytes
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stodolas

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PostPosted: Mon Jan 28, 2008 6:24 pm    Post subject:
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Really? 8 bytes? It uses less than 1/2 of its size? That doesn't seem quite logical to me... Does it seem correct to you? Which platform is that on? From what I read the size of COMP is compiler vendor specific.
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dick scherrer

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PostPosted: Mon Jan 28, 2008 10:14 pm    Post subject:
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Hello,

With Enterprise COBOL, these:

Code:
 77  SMLBINARY           PIC S9(2)  COMP  VALUE 10.   
 77  MEDBINARY           PIC S9(9)  COMP  VALUE 100.   
 77  BIGBINARY           PIC S9(18) COMP  VALUE 100.   
result in these lengths:
Code:
 BLW=00000+018         2C
 BLW=00000+020         4C
 BLW=00000+028         8C
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Shanmugavel

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PostPosted: Sat Jun 14, 2008 11:14 pm    Post subject: Reply to: COMP, COMP1, COMP2 & COMP3 in COBOL
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dick scherrer,

I didn't understand the above piece of code and BLW is.
Can you explain it some more?
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dick scherrer

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PostPosted: Sun Jun 15, 2008 2:46 am    Post subject:
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Hello,

The 77-level variables "code" is to show 3 binary numbers of different length.

The "result" is what the compiler assigned for these variables and is generated in the listing from the compile.

There are no executable instructions. These were only to demonstrate the length of these data items.
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sid_aec

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PostPosted: Tue Jul 15, 2008 3:34 pm    Post subject:
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Hi ,

I think what senario Dick actually wanted to explain is as below:

PIC S9(01) COMP -> PIC S9(04) COMP :Takes 2 bytes.
PIC S9(05) COMP -> PIC S9(09) COMP :Takes 4 bytes.
PIC S9(10) COMP or above :Takes 8 bytes.
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manishmittal

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PostPosted: Fri Jun 12, 2009 3:21 pm    Post subject:
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Hi all,

My apologies for replying to an year old post.But i just want to confirm if the above answer is correct.

Thanks.
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dbzTHEdinosauer

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PostPosted: Fri Jun 12, 2009 3:27 pm    Post subject:
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manishmittal wrote:
My apologies for replying to an year old post.But i just want to confirm if the above answer is correct.


well you could either mimic Dick's experiment, or you could check the COBOL Prg Guide and insure that sanga_1978 paraphrased everything properly.
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manishmittal

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PostPosted: Fri Jun 12, 2009 3:30 pm    Post subject:
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I tried it by defining some binary variables with different lenghts in a copubook and then opening this copybook against a dummy input file usinf Fileaid.But the results i got are conflicting with this post.
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dbzTHEdinosauer

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PostPosted: Fri Jun 12, 2009 5:10 pm    Post subject:
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manishmittal wrote:
I tried it by defining some binary variables with different lenghts in a copubook and then opening this copybook against a dummy input file usinf Fileaid.But the results i got are conflicting with this post.


what different lenghts did you use? elementary items defined as USAGE Binary are 2,4,8 bytes long, depending upon the pic clause.

what are the conflicting results?
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manishmittal

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PostPosted: Fri Jun 12, 2009 5:15 pm    Post subject:
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Please accept my apologies.After going thru the Program Reference guide and changing the copybook according to that i found the same results as mentioned in the post..

Thanks for the clarification.
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dbzTHEdinosauer

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PostPosted: Fri Jun 12, 2009 5:25 pm    Post subject:
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Manish,

apologies are not needed.
That you have learned something is of paramount importance;
that was the reason for my remarks.
good luck!
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