Re^7: Regex's, parentheses, and the mysterious ( ??{ } ) operator

"You have to properly account for all capturing groups in the overall regex..."

I did not yet work through your example program or explore the references you gave me, but your sentence above jolted my brain into some semblance of activity. Yes, it makes a lot of sense that every time the recursive regex satisfies another capture group, it uses yet another capture variable. So I printed out a bunch of them:

$ cat p7.pl 
#!/opt/perl5.16/bin/perl

use strict;
use warnings;

our $paren = qr/                # Need declared variable with use stri
+ct.
      \(
        ( 
           [^()]+               # Not parens
         | 
           (??{our $paren})     # Another balanced group (not interpol
+ated yet)
        )*
      \)
    /x;                         # 'x' means ignore whitespace, comment
+s.


my $stuff = "On the outside now then (we go( in( and in (&stop)(awhile
+) ( further ))) but still (here) ) and now ((for a while)) we are out
+ again.";
$stuff =~ /($paren)[^()]*($paren)/;

print "-original-\n";
print "$stuff\n";
print "1---------\n";
print 'X' . $1 . 'X' . "\n";
print "2---------\n";
print 'X' . $2 . 'X' . "\n";
print "3---------\n";
print 'X' . $3 . 'X' . "\n";
print "4---------\n";
print 'X' . $4 . 'X' . "\n";
print "5---------\n";
print 'X' . $5 . 'X' . "\n";
print "6---------\n";
print 'X' . $6 . 'X' . "\n";
print "----------\n";
$ ./p7.pl 
-original-
On the outside now then (we go( in( and in (&stop)(awhile) ( further )
+)) but still (here) ) and now ((for a while)) we are out again.
1---------
X(we go( in( and in (&stop)(awhile) ( further ))) but still (here) )X
2---------
X X
3---------
X((for a while))X
4---------
X(for a while)X
5---------
Use of uninitialized value $5 in concatenation (.) or string at ./p7.p
+l line 31.
XX
6---------
Use of uninitialized value $6 in concatenation (.) or string at ./p7.p
+l line 33.
XX
----------
$
[download]

So the recursive regex evaluation set $1 through $4! Thanks, this makes more sense now.

Comment on Re^7: Regex's, parentheses, and the mysterious ( ??{ } ) operator Download Code

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Re^8: Regex's, parentheses, and the mysterious ( ??{ } ) operator
by AnomalousMonk (Archbishop) on Jul 15, 2013 at 19:24 UTC

... every time the recursive regex satisfies another capture group, it uses yet another capture variable.

Old-style, numbered capture groups are counted according to the literal order in which their opening parentheses appear in the final, compiled regex; they are not created at run-time and do not depend on the evaluation, recursive or otherwise, of any regex sub-expression at run-time. (Of course, the actual capturing happens at run-time!) In the example below, the final regex $ry (after full interpolation) is printed and the capture groups marked and counted (at least, that's what I tried to do!). Sorry for any wrap-around, which may make this difficult to read. Look for examples of capture group counting in perlre and perlretut.

>perl -wMstrict -le
"my $s = 'x(y) (a(b)) ()() q (a(b)c()(d(e(f)g))h) q';
 ;;
 our $rx = qr{ \( ([^()]* | (??{ our $rx }))* \) }xms;
 ;;
 my $ry = qr{ ($rx) [^()]* ($rx) }xms;
 ;;
 $s =~ $ry;
 print qq{1st '$1'   3rd '$3'};
 ;;
 print qq{\nfinal regex:};
 print $ry;
"
1st '(y)'   3rd '(a(b))'

final regex:
(?^msx: ((?^msx: \( ([^()]* | (??{ our $rx }))* \) )) [^()]* ((?^msx: 
+\( ([^()]* | (??{ our $rx }))* \) )) )
[download]


counted capture groups:

(?^msx: ((?^msx: \( ([^()]* | (??{ our $rx }))* \) )) [^()]* ((?^msx: 
+\( ([^()]* | (??{ our $rx }))* \) )) )
        |           |                        |      |        |        
+   |                        |      |
        1st begin   2nd begin                2-end  1st end  3rd begin
+   4th begin                4-end  3rd end
[download]

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