On Jan 23, 2008 12:58 AM, Chas. Owens <[EMAIL PROTECTED]> wrote:
>
> Ah, I always get bitten by that. It needs to be
>
> print $out $_;
>
Sorry for the quick post, I figured it ou after the post.
> snip
> > Shouldn't my code keep one of the duplicate users instead of taking
> > them both out?
> snip
>
> On the first pass through for the uid 10 and uname foo, $seen{10} and
> $seen{foo} will both be one and therefore the next will not be
> triggered (it only triggers if at least one of them is greater than
> one). On the second pass through for uid 10 and uname foo, $seen{10}
> and $seen{foo} will be two and it will skip that line.
>
Yes, the algorithm has a problem, please read my answer to Tom's
e-mail and comment.
Thanks.
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