divide statement in cobol

inbasekarmani · New User Joined: 20 Apr 2007 Posts: 13 Location: chennai

hi,

WORKING-STORAGE SECTION.
77 FIELD1 PIC 9(2)V9.
77 FIELD2 PIC 9V9.

PROCEDURE DIVISION.
….
DIVIDE 5 BY 4 GIVING FIELD1 REMAINDER FIELD2.

when i execute this ,iam getting the following value

Field1 is 012
field2 is 02

when i execute the same coding with field1 pic 9(4)v9(4), then iam getting field2 as 00.

why this is happening?

Robert Sample · Posted: Mon Oct 26, 2009 5:32 pm

You need to find the manuals (link at the top of the page), pull up the COBOL Language Reference, and read up -- intensively -- on both intermediate results and COBOL division.

4 times 1<.>2 is 4.8; 5.0 minus 4.8 is 0<.>2 -- the results you got are what you should have gotten based upon your field definitions.

With a field of PIC 9(4)V9(4), the division gives 0001<.>2500 so there is no remainder so the remainder should be 00.

So why are you surprised about these results (other than you not understanding COBOL division, intermediate results and final results)?

bhairon singh rathore · New User Joined: 19 Jun 2008 Posts: 91 Location: banglore

Hi
5/4 = 1.2 as field1 can take only one decimal ...so remainder would be 0.2 which is field2

if you increase field1 to 9(4)v9(4) then value is 1.25 but then there is no remainder

Regards

Terry Heinze · Posted: Mon Oct 26, 2009 7:50 pm

Before posting your reply, did you take Robert's advice about reading about intermediate results, final result, and the DIVIDE statement in the Language Reference Manual?