COMP, COMP1, COMP2 & COMP3 in COBOL

raghu_4412 · New User Joined: 01 Mar 2004 Posts: 4 Location: Bangalore

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.

sanga_1978 · New User Joined: 07 Mar 2004 Posts: 1

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

thalapradheep · New User Joined: 05 Dec 2007 Posts: 4 Location: India

How much bytes will S9(18) Comp will occupy?

sri_mf · Active User Joined: 31 Aug 2006 Posts: 218 Location: India

How much bytes will S9(18) Comp will occupy?

8 bytes

stodolas · Posted: Mon Jan 28, 2008 6:24 pm

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.

dick scherrer · Posted: Mon Jan 28, 2008 10:14 pm

Hello,

With Enterprise COBOL, these:

Shanmugavel · New User Joined: 15 Dec 2006 Posts: 7 Location: CHENNAI

dick scherrer,

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

dick scherrer · Posted: Sun Jun 15, 2008 2:46 am

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.

sid_aec · New User Joined: 01 Jul 2008 Posts: 60 Location: Kolkata

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.

manishmittal · New User Joined: 25 Apr 2008 Posts: 49 Location: Gurgaon

Hi all,

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

Thanks.

dbzTHEdinosauer · Posted: Fri Jun 12, 2009 3:27 pm

manishmittal · New User Joined: 25 Apr 2008 Posts: 49 Location: Gurgaon

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.

dbzTHEdinosauer · Posted: Fri Jun 12, 2009 5:10 pm

manishmittal · New User Joined: 25 Apr 2008 Posts: 49 Location: Gurgaon

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.

dbzTHEdinosauer · Posted: Fri Jun 12, 2009 5:25 pm

Manish,

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