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sivaraj.ramaraj · Beginner Joined: 05 Sep 2007 Posts: 5 Topics: 3

how many bytes did S99V999 COMP-3 will take?

CraigG · Posted: Wed Aug 13, 2008 12:23 pm Post subject:

Have you tried looking in the COBOL manual? 3 Bytes.

sivaraj.ramaraj · Beginner Joined: 05 Sep 2007 Posts: 5 Topics: 3

yah , i looked into the manuls but i thought it will count 'S' also so 6/2+1=4 bytes

kolusu · Posted: Wed Aug 13, 2008 1:28 pm Post subject:

sivaraj.ramaraj,

how did you count 6 in 99V999 ? The decimal point is NOT stored

Kolusu

sivaraj.ramaraj · Beginner Joined: 05 Sep 2007 Posts: 5 Topics: 3

Hi Kolusu,
i didn't count the decimal part ,but i hve counted the 'S' bit

warp5 · Posted: Thu Aug 14, 2008 12:57 am Post subject:

S99V999 gives 6 hex characters, 5 digits, 1 sign, which in turn gives a total of 3 bytes.

sivarao · Beginner Joined: 21 Aug 2008 Posts: 6 Topics: 1

S99V999-it will occupy 3 bytes.
S99V999- 5 digits and 'V' is a suumed decimal point it will not occupy any memory.

Let me know if you have any doubt.

Santlou · Posted: Mon Sep 01, 2008 9:23 pm Post subject:

An easy way to remember is to take the number of digits (not including the sign or implied decimal), then divide that number by 2, drop the remainder (if any) and add 1.

Example:

s9(2)V999

there are 5 digits here, so 5 / 2 = 2.5
drop the remainder (.5) = 2
Add 1 (2 + 1) = 3

This is for Packed (COMP-3) fields.