Re: Re: A demanding parser

[TheDamian]

Is there any way of avoiding the external module and catch an arbitrary number of nested parentheses with a "normal" Regex?

use Regexp::Common;

$str =~ /$RE{balanced}{-parens=>'()'}/

[Juerd]

That's what he does already. He asked if it were possible without the module. (Not a very good idea, but I don't know the motivation (education, perhaps))

2;0 juerd@ouranos:~$ perl -e'undef christmas'
Segmentation fault
2;139 juerd@ouranos:~$

[TheDamian]

Ah, sorry. Missed that bit.

How about:

    use Regexp::Common;

    print $RE{balanced}{-parens=>'()'}, "\n";
    print $RE{balanced}{-parens=>'{}'}, "\n";
    print $RE{balanced}{-parens=>'{}()'}, "\n";
    # etc.

followed by reading the Regexp::Common source?

Damian

[gmax]

It works! I had some trouble figuring out how to implement it, because with 'use strict' the recursive RegEx stays undefined.


#!/usr/bin/perl -w
use strict;
my $input = "aa (bb cc) dd (ee (ff (gg) hh) jj) kk ((ll)) mm";
no strict 'vars';
# from $RE{balanced}{-parens=>'()'};
$REparens = qr/
    (?:(?:(?:[(](?:(?>[^)(]+)
    |(??{$REparens}))*[)]))
    |(?:(?!)))/x;

my $re = $REparens;    
use strict;

print "input: $input\n";
while ($input =~ /\G(\w+|$re)\s*/gc) {
    print "$1\n"
}
__END__
output:
aa
(bb cc)
dd
(ee (ff (gg) hh) jj)
kk
((ll))
mm

Making a block of no strict 'vars' solves the problem. There I can assign the recursive regex to a proper ly declared variable, or call explicitly ($main::REparens).
Thanks a million.


 _  _ _  _  
(_|| | |(_|><
 _|