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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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