Re: Half-Adjust with /free

Thanks Scott -

I wonder if this would work:

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

       /free
          a = %dech( b + c : %len(a): %decpos(a));

And that way it would be more or less dynamic. 

Hmmm...doesn't seem to work...gives me a compile error. Odd though...it
knows the %len and the %decpos of variable a at compile time. 

- Michael


On Tue, 22 Jun 2004 10:57:59 -0500 (CDT), "Scott Klement"
<rpg400-l@xxxxxxxxxxxxxxxx> said:
> 
> 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
> 
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