... 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]

In reply to Re^8: 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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