Reg :storage allocation of Comp variables(from IBm manual).

prabhucs01 · New User Joined: 07 Mar 2005 Posts: 40

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) | |__________________________________|_________________________________|

i just found this storage allocation chart from the IBM manual . Could anyone tell me how a single digit in a picture clause gets allocated.

For example PIC 9 gets its storage allocation. Does it take 4 bits for one Digit in picture class?

If so, then total Nine digits [pic 9(9) ] will take 36 bits (not 32 bits or full word)

Pardon me if this is an repeating question.

agkshirsagar · Posted: Mon Jul 21, 2008 3:40 pm Post subject:

I think you are misinterpreting the manual.
It clearly says for 1 to 4 digits it will occupy 2 Bytes.
PIC 9 will need 2 bytes of storage.
Not like you are thinking 1 digit 4 bits and so on..

UmeySan · Posted: Tue Jul 22, 2008 4:53 pm Post subject:

Hi !

For informations of storage allocations of comp fields, plse use the
search function. This has been discussed till death.

prabhucs01 · New User Joined: 07 Mar 2005 Posts: 40

ok...sorry mates