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Shannon ODonnell wrote:
Given the following code fragment:


D ImgCnt S 3 0 D ImgCntc S 3a C Do 999
C Eval ImgCnt = ImgCnt + 1 C Eval ImgCntc = %Char(ImgCnt) C Enddo

Or you could do this:
D DS
D ImgCnt 1 3S 0
D ImgCntc 1 3A

and then eliminate the "ImgCntc = %Char(ImgCnt)" entirely.


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