Re: getting a return code from a looping regular expression

[splinky]

Utilizing Perl 5.6 regex features, you can do the following:

our $parens;
my  $re = qr{ (
              \( (?{$parens++ if $parens >= 0})
              |
              \) (?{$parens-- if $parens >= 0})
              |
              .
              )*
            }x;

my $this = 'a(bc(de)fg)h';
my $that = '(a(bc(de)fg)h';
my $other = 'a(bc(de)fg)h)';
my $bla = ')a(bc(de)fg)h)';

$parens = 0;
print "this, $parens\n" if $this =~ /$re/;
$parens = 0;
print "that, $parens\n" if $that =~ /$re/;
$parens = 0;
print "other, $parens\n" if $other =~ /$re/;
$parens = 0;
print "bla, $parens\n" if $bla =~ /$re/;

Which prints:

this, 0
that, 1
other, -1
bla, -1

Specifically, -1 indicates that, at some point, an attempt was made to close parens that hadn't been opened. A positive number indicates how many unclosed left parens were encountered.

Thanks, Ilya.

*Woof*