Re: How to remove the $1 hard coding

You could do

{
   no strict 'refs';
   $somestring =~ m/\s*(\S+)\s*(\S+)\s*(\S+)\s*(\S+)\s*/;
   $result = ${$matchnum};
}
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but it is strongly discouraged (forget about it immediately), and won't even work under strict (that's the reason for the no strict 'refs';). Fletch's answer provides a much better solution, to which I could just add that you can avoid copying the matches to an array, if you treat the return values of the match as a list:

push @sps, (m/\s*(\S+)\s*(\S+)\s*(\S+)\s*(\S+)\s*/)[$columnNumber];
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should work fine.
Cheers, CombatSquirrel.

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Re: Re: How to remove the $1 hard coding by hardburn (Abbot) on Aug 22, 2003 at 16:44 UTC
Actually, after thinking about it, I don't think a symbolic ref solution would be so bad in this case, provided you take a few percautions. At the top of your subroutine, make sure `$matchnum` matches against `/\A\d+\z/` and that `$$matchnum` is a defined value after the match (change your last line to `$result = $$matchnum if $$matchnum;`). Though overall, you're still probably better off with the array solution. ---- I wanted to explore how Perl's closures can be manipulated, and ended up creating an object system by accident. -- Schemer Note: All code is untested, unless otherwise stated	[reply] [d/l] [select]
Re: Re: Re: How to remove the $1 hard coding by shenme (Priest) on Aug 22, 2003 at 21:26 UTC
You meant, of course, `$result = $$matchnum if defined $$matchnum;`. And while I agree the array approach is better, your mention of `$$matchnum` "made me look". I hadn't thought the `$1` variables were _that_ global.	[reply]