in reply to Re^2: Regex for matching n-fold repititions of arbitrary characters
in thread Regex for matching n-fold repititions of arbitrary characters

Depends on what you want /./ to match. If you want to match repetitions of any character, yes. If you want to match repetitions of specific characters, no, since you won't be using /./.

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Re^4: Regex for matching n-fold repititions of arbitrary characters
by pat_mc (Pilgrim) on Oct 08, 2009 at 19:49 UTC
    Hm ... I follow you so far, ikegami ... but why do I get different output in the following two cases:
    ~$ perl -le 'print $& if "aabbaab" =~ /aa|bb/g'

~$ perl -le 'print join "\n", ( "aabbaab" =~ /aa|bb/g )'
aa
bb
aa

~$ perl -le 'print join "\n", ( "aabbaab" =~ /(.)\1/g )'
a
b
a

    [download]

    The first two system responses are clear ... but why does the last regex using backreferences return only single characters?
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[d/l]

      /pat/g means /(pat)/g if pat contains no captures. In list context, m// returns what it captured (or 1 if there are no captures) on success. 

      $ perl -wle'print for "aabbaab" =~ /((.)\2+)/sg;'
aa
a
bb
b
aa
a

      [download]
      $ perl -wle'print $1 while "aabbaab" =~ /((.)\2+)/sg;'
aa
bb
aa

      [download]

      Note that if "aabbaab" =~ /aa|bb/g is wrong. Red flags should be raised when you see /g in scalar context outside a loop.

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[d/l]
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        Note that if "aabbaab" =~ /aa|bb/g is wrong. Red flags should be raised when you see /g in scalar context outside a loop.

        Not necessarily! It makes sense when parsing with nested ifs! Friedl's book has some nifty examples...

        Cheers Rolf

[reply]

      Because you are only capturing one character. m// in list context returns the list of captures, which in your case is the (.). The backreference is not captured.

      print pack("A25",pack("V*",map{1919242272+$_}(34481450,-49737472,6228,0,-285028276,6979,-1380265972)))
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[d/l]
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      maybe that's clearer:

      $ perl -e 'print $1,$2,"\n" while ( "aabbaab" =~ /(.)(\1)/g )'
aa
bb
aa

      [download]
      or 
      $ perl -e 'print $&,"\n" while ( "aabbaab" =~ /(.)\1/g )'
aa
bb
aa

      [download]

      Cheers Rolf

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[d/l]
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