Re: Eternal question of parsing parentheses

Eternal answer RTFMš ! :)

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

(š) SCNR 8)

UPDATE: untabified code.

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Re^2: Eternal question of parsing parentheses by LanX (Saint) on Oct 24, 2009 at 20:13 UTC
Hi I simplified the code, and replaced tr/()/<>/ to make it more readable: `$_='a<b<c<d><e>>f>g<h><<i>j>'; $re = qr{ < # anchor at first paren as wanted ( # paren group 1 (?: (?> [^<>]+) # Non-parens without backtracking \| < (?1) # Recurse to start of paren group 1 > )* ) }x; /$re/; print $1;` [download] `perl /tmp/tst2.pl b<c<d><e>>f` [download] Cheers Rolf	[reply] [d/l] [select]
Re^2: Eternal question of parsing parentheses by AnomalousMonk (Archbishop) on Oct 24, 2009 at 18:46 UTC
For those still limited to 5.8 and before, see also the discussion of the `(??{ code })` construct under Extended Patterns (just before the discussion of `(?PARNO)`) in perlre.	[reply] [d/l] [select]