Eternal answer RTFM¹ ! :)

perldoc perlre ( >5.10) and searching for "recursive" brought the following code, where I only had to exchange "foo" with \w*

$_='a(b(c(d)(e))f)g(h)((i)j)';
 
$re = qr{ (                     # paren group 1 (full function)
            \w*
            (                   # paren group 2 (parens)
              \(
              (                 # paren group 3 (contents of parens)
                (?:
                  (?> [^()]+ )  # Non-parens without backtracking
                |
                  (?2)          # Recurse to start of paren group 2
                )*
              )
              \)
            )
          )
      }x;
 
@matches=/$re/;
print "@matches";

[download]

perl /tmp/tst.pl
a(b(c(d)(e))f) (b(c(d)(e))f) b(c(d)(e))f
Compilation finished at Sat Oct 24 19:42:58

[download]

Cheers Rolf

(¹) SCNR 8)

UPDATE: untabified code.

In reply to Re: Eternal question of parsing parentheses by LanX
in thread Eternal question of parsing parentheses by Anonymous Monk

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