in reply to Re: getting a return code from a looping regular expression
in thread getting a return code from a looping regular expression

while s/[(][^()][)]//; is actually a syntax error. Also, that pattern only matches parenthesis with exactly one character between them.

Here's some working code for arbitrary parenthesized strings:

sub balanced
{
        $_ = shift;
        while (s/[(][^()]*[)]//) {};
        return 0 if /[()]/;
        return 1;
}

[download]

Of course a real programmer uses a PDA to balance parenthesis. (;

sub balanced
{
        my $string = shift;
        my $stack  = 0;

        for my $index (0..length($string) - 1)
        {
                my $char = substr($string, $index, 1);
                if ($char eq '(')
                {
                        $stack++;
                } elsif ($char eq ')') {
                        $stack--;
                }

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                # Update: Ha!  Thanks, BlaisePascal.
                # A PDA that happily pops an empty
                # stack isn't very useful. :)
                return 0 if $stack < 0;
    }

        return $stack ? 0 : 1;
}

[download]

-Matt

Replies are listed 'Best First'.
RE: RE: Re: getting a return code from a looping regular expression
by BlaisePascal (Monk) on Aug 01, 2000 at 22:57 UTC
    True, PDA's are the proper tool for balancing parenthesis. But it helps if it works... Your PDA would match "))((" perfectly find. Most wouldn't consider that balanced... 
    sub balanced
{
  my $string = shift;
  my $stack = 0;

  @string = split //, $string;
  for (@string) {
      last if $stack < 0;
      $stack++ if /[(]/;
      $stack-- if /[)]/;
  }
  return $stack ? 0 : 1;
}

    [download]
    or, without using split and //: 
    sub balanced
{
  my $string = shift;
  my $stack = 0;

  for my $i (0..length($string)-1) {
    last if $stack < 0;
    $stack++ if substr($string,$i,1) eq '(';
    $stack-- if substr($string,$i,1) eq ')';
  }
  return $stack ?0 : 1;
}

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