Just another way to do it without loops (kind of cheating):
my $c=0; s/ \( (?{++$c}) | \) (?{--$c}) | ([^()]+) /$1 if ($c==0)/gex;
If you match an open paren, increment a counter;
if you match a close paren, decrement it;
if you match a string of non-parens, capture it;
substitute whatever was captured back in if the counter is zero.

There is an "if", but it can be replaced by an "x" operator, if you're a stickler. Or you could rewrite the replacement as $c==0 and $1.

Another little variation that doesn't rely on the (?{}) construct:

s{([()]*)([^()]+)} {$c += $1=~y/(// - $1=~y/)//; $2 x !$c }ge;
Match any parens followed by any non-parens;
Add the count of opens and subtract the count of closes from your counter;
Substitute the non-paren portion back in if the counter is zero.

We're not really tightening our belts, it just feels that way because we're getting fatter.

In reply to Re: regular expression parentheses remover by Roy Johnson
in thread regular expression paranthesis remover by Kanishka

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