adrianh has asked for the wisdom of the Perl Monks concerning the following question:

Is there a nice way I can figure out whether my regexp has capturing matches in it? My question turned up with some code that (boiled down) needs to differentiate between these two cases:

my @x = ( "1234" =~ m/(\d)/ );  # @x contains a captured match of "1"
my @y = ( "1234" =~ m/34/ );    # @y contains a "1" for a successful m
+atch
print "x=(@x), y=(@y)\n";
# x=(1), y=(1)

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Now I guess I could set $1 to undef before the match and test it afterwards - but this seems so darn inelegant. Am I missing a trick?

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Re: Determing whether a regexp has capturing matches
by rhesa (Vicar) on Feb 27, 2007 at 15:52 UTC
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      D'oh. Of course. Thanks. 
      sub has_captured_matches { return $#+ };

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        Keep in mind that $#+ is -1 when the pattern does not match, regardless of whether there are captures. 

        "1234" =~ /(1)34/;
print $#+, "\n";
"1234" =~ /134/;
print $#+, "\n";

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        That prints "-1" both times, and -1 is considered a "true" value.

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Re: Determing whether a regexp has capturing matches
by kyle (Abbot) on Feb 27, 2007 at 15:45 UTC

    You don't need to set $1 before the match. In fact, I think you can't. 

    $ perl -le '$1=undef;'
Modification of a read-only value attempted at -e line 1.

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    $1 will be set to undef after a match with no parentheses. 

    "1234"=~/(\d)/;
print "$1\n";
"1234"=~/34/;
print "$1\n";
print "undef\n" if ! defined $1;

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    Prints: 

    1

undef

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    This doesn't work when a regex had capturing parentheses that didn't match. Just like the "no parentheses" case above, this also prints "undef": 

    "1234"=~/(x)?34/;
print "undef" if ! defined $1;

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      $1 will be set to undef after a match with no parentheses.

      Is this documented behaviour? I could have sworn I'd used a perl at some point where $1 persisted unless you made another captured match. I could, of course, be making shit up :-)

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