paulh at harlequin
Sep 26, 1995, 2:07 PM
Post #1 of 1
(94 views)
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PM> 2. ($out_buffer, $status) = doit($in_buffer) ;
I like ($status, $outbuffer) = ....
It makes it clear the $outbuffer is a result, not an input, and makes
it slightly more difficult to ignore status than your original version
(since you have to put something to get the status value...)
P.
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