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

Dear Monks 
     if(/[^\d\D]/)
     { 
          print "match" 
     }

[download]

it is getting match dot "." character how ... ?

Thanks

Replies are listed 'Best First'.
Re: Pattern Matching
by grizzley (Chaplain) on Aug 13, 2009 at 10:27 UTC
    No it doesn't. At least on my machine... :)
    /[^\d\D]/ in my opinion will never match.
[reply]
[d/l]
      I think you're right :) 
      printf "(%s) match (%d)\n",$_, 0+/[^\d\D]/
    for qw[ !! aa AA 11 99 )( :D .. ];
__END__
(!!) match (0)
(aa) match (0)
(AA) match (0)
(11) match (0)
(99) match (0)
()() match (0)
(:D) match (0)
(..) match (0)

      [download]
[reply]
[d/l]
Re: Pattern Matching
by Anonymous Monk on Aug 13, 2009 at 08:53 UTC 
    use YAPE::Regex::Explain;
print YAPE::Regex::Explain->new(qr/[^\d\D]/)->explain;
__END__
The regular expression:

(?-imsx:[^\d\D])

matches as follows:

NODE                     EXPLANATION
----------------------------------------------------------------------
(?-imsx:                 group, but do not capture (case-sensitive)
                         (with ^ and $ matching normally) (with . not
                         matching \n) (matching whitespace and #
                         normally):
----------------------------------------------------------------------
  [^\d\D]                  any character except: digits (0-9), non-
                           digits (all but 0-9)
----------------------------------------------------------------------
)                        end of grouping
----------------------------------------------------------------------

    [download]
    any character except digits and non-digits is every character.
[reply]
[d/l]
      No. Any character except digits or non-digits would be every character. With and, this regexp will never match.
[reply]
        I think i was trying to say: Any charater except every character.
[reply]

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