|
[SNIP]
Code that works:
> > wrk = (%dec(transNo:%size(transNo):0) + 1);
> > parms.transNo = %editc(wrk:'X');
[SNIP]
Code that fails:
> > parms.transNo = %editc((%dec(transNo:%size(transNo):0) + 1):'X');
Hi,
As virtually everyone has already said, the problem here is the
intermediate representation.
The %dec() BIF is outputting a 10,0 number. Since you then proceed to
add 1 to it, it adds an additional digit to handle the largest possible
value that could result.
What people didn't tell you (or at least I didn't see it mentioned) is
that you can use %dec() BIF to control the size of the intermediate
representation. Simply wrap your arithmetic expression in an additional
%dec() BIF and specify 10,0... this will cause the expression to use a
10,0 intermediate...
Of course, the resulting code is pretty ugly :) I've indented it to make
it a bit more obvious what I'm doing:
// original expression:
parms.transNo = %editc(
(%dec(transNo: %size(transNo): 0) + 1)
: 'X');
// controlling intermediate results with %dec()
parms.transNo = %editc(
%dec(
%dec(transNo: %size(transNo): 0) + 1
: %size(parms.transNo): 0)
: 'X');
HTH
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