... $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))'
--------

[download]

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

In reply to Re^6: Regex's, parentheses, and the mysterious ( ??{ } ) operator by AnomalousMonk
in thread Regex's, parentheses, and the mysterious ( ??{ } ) operator by Clovis_Sangrail

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