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)(.*)//;

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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';
}

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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!

In reply to Re: regx by swiftone
in thread regex not working by jumk

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