in reply to Re^2: Matching nested parens
in thread Matching nested parens

A footnote. I was curious about Regexp::Common. So: 
use Regexp::Common qw/balanced/;
say $RE{balanced}{-parens=>'()'} ;

[download]
And I can now see why I didn't have a hope of figuring out the RE on my own... :o)

Replies are listed 'Best First'.
Re^4: Matching nested parens
by AnomalousMonk (Archbishop) on Oct 01, 2020 at 23:24 UTC
    ... I didn't have a hope of figuring out the RE on my own...

    This problem is actually used as an an example of the use of (?R) and friends in perlre: 

    Win8 Strawberry 5.30.3.1 (64) Thu 10/01/2020 19:06:47
C:\@Work\Perl\monks >perl -Mstrict -Mwarnings
use 5.010; # need 5.10+ extended patterns (?R)

use Data::Dump qw(dd);

my $str = '(a a (a a)(a a))(b b(b)b b) xxx (c (c (c c)))';

my $rx_bal_parens = qr{
    \(
      (                # Capture group 1 (contents of parens)
        (?:
          (?> [^()]+ ) # Non-parens without backtracking
        |
          (?R)         # Recurse to start of regex
        )*
      )
    \)
    }xms;

my @bal_parens = $str =~ m{ $rx_bal_parens }xmsg;

dd \@bal_parens;
^Z
["a a (a a)(a a)", "b b(b)b b", "c (c (c c))"]

    [download]
    (Of course, if you use something like this, do thorough testing to be sure it really does what you want. :)

    Update: Fixed first line of code (the use 5.010; statement). It somehow got stuck onto the end of the command line during cut-pasting.

    Give a man a fish:  <%-{-{-{-<

[reply]
[d/l]
[select]
      I can see that I'm *WAY* behind the curve. I thought I about the various (? RE operators but I've heard of the ?R. Whew. Recursion inside a regular expression... I wonder if Perl's REs are Turing complete :o) Thanks!
[reply]

In Section Seekers of Perl Wisdom