Thanks, Terry. Except for the need to code an indeterminate number of
zeros in that string (though, in reality, I can assume that no purely
numeric column will be wider than, say, 15 positions), I think your
solution is more elegant, meaning your 1 instruction rather than my 3
(since I already needed an 'if len_this_col < Max_len_this_col to
condition my FOR loop.)
Arthur J. Marino
RockTenn Corporation
(631) 297-2276
-----Original Message-----
From: rpg400-l-bounces@xxxxxxxxxxxx
[mailto:rpg400-l-bounces@xxxxxxxxxxxx] On Behalf Of Terrence Enger
Sent: Wednesday, March 25, 2009 1:18 PM
To: RPG programming on the IBM i / System i
Subject: Re: Using an Expression to control a FOR loop
Arthur,
If I understand what you are doing, you could say something like ...
if %len( coldta ) < cwdth( #col );
coldta = %subst( '000000000000000' // "lots" of zeros
: 1
: cwdth( #col ) - %len( coldta )
) + coldta;
endif;
Whether this is easier to understand than an explicit loop, is a matter
of taste. I confess that I find neither way very transparent. Can
someone here give us a clearer solution?
Cheers,
Terry.
On Wed, 2009-03-25 at 11:01 -0400, Arthur Marino wrote:
This FOR loop is supposed to prefix the 'varying' character field
"coldta" with zeros so as to right-justify the value in the column's
defined field length. "i" is defined as 5i 0.
for i = 1 to (cwdth(#col)-%len(coldta));
coldta = '0' + coldta;
endfor;
Debug tells me:
cwdth(#col) = 5 (the column's defined field length)
coldta = '134' (%len(coldta)=3, I inserted an explicit calc to be
sure)
"coldta" should wind up being '00134'.
I expect the loop to iterate twice (5 - 3). Yet, it does so only ONCE.
After the 'endfor' i = 2, not 3 as I'd expect, and coldta contains
0134
(only 4 bytes).
When I insert the calc "len = cwdth(#col)-%len(coldta);" and replace
the
expression with "for i = 1 to len", it iterates twice, i = 3 after the
'endfor', and coldta=00134 (5 bytes).
Is there something I'm not getting about the evaluation of the
expression in the FOR loop? It certainly appears that 5 - 3 results in
1, not 2.
Thanks.
Arthur Marino
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