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Re: regx

by swiftone (Curate)
on Jan 24, 2003 at 00:09 UTC ( #229490=note: print w/ replies, xml ) Need Help??


in reply to regex not working

This has been answered , but I'll explain some of the details:

The s/// command returns either the number of replacements ( in scalar context (i.e. one value)), or the values of any captured patterns in list context. (e.g. ($foo, $bar) = s/(my)(.*)// will assign "my" to $foo and anything else to $bar)

By default s/// operates (performs the search and replace) on $_. You can bind another variable to be search-and-replaced with =~. Thus, the following is perfectly valid code:

$baz = "mythingie"; ($foo, $bar) = $baz =~ s/(my)(.*)//;
and will result in $foo = 'my', $bar = 'thingie', $baz=''; Note that $foo and $bar will be undefined if the search pattern does not match. It's a common mistake to assume the pattern works. It's probably better to have code like:
use strict; $baz = 'mythingie'; if($baz =~ s/(my)(.*)//){ #s/// returns '' (which is false) if it doesn't match. #No, I don't know why it doesn't return 0. $foo = $1; $bar = $2; } else { warn 'I thought this would always match, but I must be wrong'; }
The same general rules apply to m//, though of course there is no replacement, and the bound variable is not altered.

All of this is covered in perlop, though it's buried in there a ways.

Hope that helps!


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