Re^4: Regex's, parentheses, and the mysterious ( ??{ } ) operator

[Clovis_Sangrail]

Ahh! I have deciphered the meaning of your koan. I get a compilation error when I take the "our" out of the middle of the recursive regex:

$ vi p6.pl
"p6.pl" 36 lines, 925 characters 
#!/opt/perl5.16/bin/perl

use strict;
use warnings;

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


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

print "----------\n";
print "$stuff\n";
print "----------\n";
"p6.pl" 27 lines, 650 characters 
$ ./p6.pl 
Variable "$paren" is not imported at (re_eval 1) line 2.
Global symbol "$paren" requires explicit package name at (re_eval 1) l
+ine 2.
Compilation failed in regexp at ./p6.pl line 14.
$

[Clovis_Sangrail]

If I delete the "use strict;" the program compiles/runs, but still does not match the 2nd regex. I also tried making a duplicate of $paren called $par2, and used it instead of $paren to set $2, and that doesn't work either, $2 is still blank.

[AnomalousMonk]

... $2 is still blank.

You have to properly account for all capturing groups in the overall regex, properly counting capture groups in any interpolated regex. If you have a capturing group within the interpolated recursive regex (not, IMHO, necessary), then you want to access the 3rd capture group variable. However, a capture group in the recursive regex screws up the more general
    @p = $s =~ m{ $r4 }xmsg;
extraction regex. See examples below.

>perl -wMstrict -le
"my $s = 'x(y) (a(b)) ()() q (a(b)c()(d(e(f)g))h) q';
 ;;
 our $r3 = qr{
   \(
     (?:
       [^()]+
       |
       (??{ our $r3 })
     )*
   \)
   }xms;
 ;;
 my @p = $s =~ m{ $r3 }xmsg;
 print qq{'$_'} for @p;
 print '--------';
 ;;
 $s =~ m{ ($r3) [^()]* ($r3) }xms;
 print qq{1 '$1'   2 '$2'};
 print '--------';
 ;;
 our $r4 = qr{
   \(
     (
       [^()]*
       |
       (??{ our $r4 })
     )*
   \)
   }xms;
 ;;
 @p = $s =~ m{ $r4 }xmsg;
 print qq{'$_'} for @p;
 print '--------';
 ;;
 $s =~ m{ ($r4) [^()]* ($r4) }xms;
 print qq{1 '$1'   3 '$3'};
 print '--------';
"
'(y)'
'(a(b))'
'()'
'()'
'(a(b)c()(d(e(f)g))h)'
--------
1 '(y)'   2 '(a(b))'
--------
''
''
''
''
''
--------
1 '(y)'   3 '(a(b))'
--------

Have you tried the very neat  "(?PARNO)" operator available with Perl 5.10+ and discussed in the example referred to here?

[Clovis_Sangrail]

"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
----------
$

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