in reply to Re: Leaking Regex Captures
in thread Leaking Regex Captures

I have been trying to understand how SuicideJunkie's code causes the results it does, and I am getting lost.
jwkrahn - do you mean that $1 etc... are not being reset if they do not match? So as SuicideJunkie asked - how come they seem to inherit the value of the 'next' match? i.e. $2 = $3 (but only if $3 matches first...)?

I also found that the \s* part of the regex is causing some of the problem - i.e. see regex 2 below - in isolation it works as expected),
but then i moved around the order of the regexes and came back the same problem with the 'fixed' regex 2 now 'inheriting' the faulty results from regex 1.

use strict;

####################
# in summary:
#  - regex 3 (same as originally posted more or less) 
#             gives the 'wrong' results
#  - regex 2 (eq regex 2 but wothout the \s* part before the letter ma
+tch) 
#             gives the 'right' results
#  - regex 1 (same as 3)
#             if excecuted before regex 2, screws the results from reg
+ex 2 as well!
# Aaaargh!
####################

print "Enter your test strings:\n";
while (<main::DATA>)
{
    chomp;
    print "\n***Testing '$_':\n";

###########
# NB: regex 1 eq regex 3
# includes \s*
# if removed 2 and 3 appear to work correctly
###########

m/
      ^(?:
        (?:
          (\d+)
          \s*    ## the offending bit
              a\s*
        )
      |
        (?:
          (\d+)
          \s*b\s*
        )
      |
        (?:
          (\d+)
          \s*c\s*
        )
      )+/ixg;

    print "Capturing \\d+ only (with \\s\*):  1='$1', 2='$2', 3='$3'\n
+";

###############
# regex 2
# no spaces
# gives expected output ONLY if regex 1 is removed
###############

    m/
      ^(?:
        (?:
          (\d+)
          a\s*
        )
      |
        (?:
          (\d+)
          b\s*
        )
      |
        (?:
          (\d+)
          c\s*
        )
      )+/ixg;

    print "Capturing \\d+ only:             1='$1', 2='$2', 3='$3'\n";
##############
# regex 3
# gives correct output only if only preceeded by regex 2
##############
m/
      ^(?:
        (?:
          (\d+)
          \s*a\s*
        )
      |
        (?:
          (\d+)
          \s*b\s*
        )
      |
        (?:
          (\d+)
          \s*c\s*
        )
      )+/ixg;

    print "Capturing \\d+ only (with \\s\*):  1='$1', 2='$2', 3='$3'\n
+";


    /^(?:(?:(\d+\s*a)\s*)|(?:(\d+\s*b)\s*)|(?:(\d+\s*c)\s*))+/ix;
    print "Capturing \\d+ plus the letter:  1='$1', 2='$2', 3='$3'\n";
}

__DATA__
1a
2b
3c
2a3b
1b1b
1b2c
2c1a
1a2c
1c2a
1a2b3c
3c2b1a

[download]

This behaviour really confuses me! And sorry to SuidiceJunkie again for jumping on his node!!

Just a something something...

In Section Seekers of Perl Wisdom