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[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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