Re: Newbie Question

theHip<the_tragic_hip@xxxxxxxxx> 05/31/06 9:52 AM >>>
What is stored is 2, but it is stored in binary. This seems like a
weird thing to do, truncated a value but stored as binary. How is
VAR1 used?

What version of CoBOL are you using? Compiler and OS.

I thought that that was what was stored there, but I couldn't make
sense of it. This routine is a check to make sure an account number is
valid. It goes through a whole bunch of multipys and divides and such,
and then the outcome of all this must match var2.

Unfortunately, all I was handed was a piece of paper with the code, no
clue what version of CoBOL was being used or compiler. I do have an
exe of the routine that I can run on XP.

Let me see if I have this part correct.

01 JB-WORK
03 VAR1 PIC 9(10)
03 VAR1-R REDEFINES VAR1
05 VAR9 PIC9(9).
05 VARLST PIC 9.

I am assuming that if 138755482 is moved to var1 then var9 would be
138755482 ... is that correct?

Thanks so much for the quick reply.

Sounds like a mod-10 or mod-11 check, with the last digit of the number
being the check-digit, so with the above var9 would be 123785548 and varlst
(the check digit) would be 2.

Check out the following:
for mod-11: http://www.augustana.ca/~mohrj/algorithms/checkdigit.html#isbn
for mod-10: http://www.augustana.ca/~mohrj/algorithms/checkdigit.html#ibm

I think each has variants, so you'll have to look at the code, but this
might help explain it.

Frank

---
Frank Swarbrick
Senior Developer/Analyst - Mainframe Applications
FirstBank Data Corporation - Lakewood, CO USA
.

Relevant Pages

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