Re: Half-Adjust with /free

Hi Michael,

> I can't believe I haven't encoutered this yet. How does one perform
> half-adjust when using free format?
>   a =(H) b + c;
> doesn't seem to work. Do I need to do a %DECH or something?

You can do it either way.  The following code works nicely:

     D a               s              9P 2
     D b               s              5P 0 inz(1232)
     D c               s              8P 3 inz(654.345)

      /free
         eval(h)  a = b + c;

But, having to include the word "eval" sometimes throws off your code...
especially if you have a bunch of calcs together and then suddenly one of
them has an "eval" in front of it.  So, you can also use %dech()

      a = %dech( b + c : 9: 2);

%dech() can be awkward because you have to specify the number of digits &
decimal places -- plus, I've always felt that hard-coding that is not a
good idea.

An alternative to hard-coding would be to use a named constant, as
follows:

     D b               s              5P 0 inz(1232)
     D c               s              8P 3 inz(654.345)

     D a               s              9P 2
     D a_digits        c                   %len(a)
     D a_decpos        c                   %decpos(a)

      /free
         a = %dech( b + c : a_digits: a_decpos);

HTH