How much bytes do S9(7)V99 COMP-3 occupy?

guruji · New User Joined: 13 Apr 2007 Posts: 61 Location: Chennai

Hi,

Could you please tell me How much bytes do S9(7)V99 COMP-3 occupy?
could you please explain the same?

Thanx,
reni

radha reddy · New User Joined: 15 May 2008 Posts: 9 Location: hyderabad

It occupies 5 bytes.

no of 9's = 9
(9+1)/2 = 5

dick scherrer · Posted: Tue Jul 08, 2008 1:08 pm

Hello,

5 bytes.

shakawa · New User Joined: 11 Jun 2008 Posts: 1 Location: Beijiing

Hi dick,

So you mean the sign also occupys one byte, do you?

dbzTHEdinosauer · Posted: Tue Jul 08, 2008 3:47 pm

Dick,

it would be so much easier for the OP if you would just teach him what he needs to learn. Apparently he can web-browse only this site - thus no access to a site like IBM, where he could actually learn something on his own. he could learn something here if he would search... but, again, easier to ask than look.

dick scherrer · Posted: Tue Jul 08, 2008 7:50 pm

Hello shakawa and welcome to the forums,

surya.kalyan · New User Joined: 09 Jan 2007 Posts: 20 Location: Mumbai

Here is a short representation of how data is stored in comp-3 fields.

ASSUME THE PICTURE CLAUSE IS 9(4) COMP-3. Number of bytes = (4+1)/2 = 3 (rounded).

So, At Max 4 digits can be accomodated.

If the number is +1234.It is stored as,
(In Hex) 01 23 4C

If the number is -1234.It is stored as,
(In Hex) 01 23 4D

C for + sign and D for -sign.

If the number is +123. It is stored as
(In Hex) 00 12 3C

If the number is -123. It is stored as
(In Hex) 00 12 3D.

One byte will store two digits. And the sign is stored in the last nibble.
We cannot allocate nibbles(half byte).

So, thats the reason for the formula
(n+1)/2.

Suggest or correct me if I am wrong somewhere.

hanucob · New User Joined: 15 Jul 2008 Posts: 3 Location: bangalore

The formula for it is INT(N/2+1)

radha reddy · New User Joined: 15 May 2008 Posts: 9 Location: hyderabad

Hi Hanucob,

There is correction to your formula.

For comp-3 formula is (N+1)/2
Sign byte occupies only 1/2 byte.

Thanks,
Radha

surya.kalyan · New User Joined: 09 Jan 2007 Posts: 20 Location: Mumbai

Hi Radha

I feel both of them are correct.

dbzTHEdinosauer · Posted: Wed Jul 16, 2008 1:02 pm

surya.kalyan,

radha reddy · New User Joined: 15 May 2008 Posts: 9 Location: hyderabad

Hi Surya,

For S9(7)V99 COMP-3

When we use N/2+1 we will have 9/2 = 4.5
4.5 + 1 = 5.5
rounding off will give 6 bytes

When we use formula (N + 1)/2
We will get (9+1)/2 = 10/2 = 5 bytes.

So (N+1)/2 or (N/2 +1/2) is the correct formula.

Correct me if I am wrong.

Thanks,
Radha

surya.kalyan · New User Joined: 09 Jan 2007 Posts: 20 Location: Mumbai

surya.kalyan · New User Joined: 09 Jan 2007 Posts: 20 Location: Mumbai

Hi radha,

ashimer · Posted: Wed Jul 16, 2008 9:16 pm

You can use both ... if you are using N/2 + 1 then take only the integer part and if you are using (N+1) / 2 then roundoff ...

use both on S9(7)v99 and S9(10)v99 and see ..... physically sign occupies only half a byte ...

Tom Goodhart · New User Joined: 16 Jul 2008 Posts: 5 Location: Tennessee

Not coding the S for sign in the picture will cause values to be stored with "F" as the sign. An positive integer value which mathematically equal to +1.

If you code an even number of digits, the compiler will always make it 1 byte larger internally.
Thus,
S9(6) COMP-3 will take 4 bytes of storage
S9(7) COMP-3 will take 4 bytes of storage This is the preferred way as it is clear to anyone what was intended.

Gousiya Mulla · New User Joined: 02 Jun 2008 Posts: 87 Location: Bangalore

Robert Sample · Posted: Fri Jul 18, 2008 8:30 pm

No:

Hex 0004D2 is the BINARY (or COBOL COMP) value of +1234.
Hex 01234C is the PACKED-DECIMAL (or COBOL COMP-3) value of +1234.
Hex F1F2F3F4 is the unsigned COBOL DISPLAY value of 1234.
Hex F1F2F3C4 is the signed COBOL DISPLAY value of +1234.

Gousiya Mulla · New User Joined: 02 Jun 2008 Posts: 87 Location: Bangalore

Robert ,

Understood .. Thax