Re^3: How to find the outermost pair of square brackets with a regex?

That code is broken.

You can't declare $re on the same line as you're using it. It won't even compile (under strict vars).
There's a I reason I used a package variable instead of a lexical. It'll bite you if you interpolate into $re. Best to always use a package variable to avoid the problem entirely.
While not a bug per say, the s modifiers on both regeps are useless because you don't use ..

        our $re
        local $re = qr{\[(?:(?>[^\[\]]+)|(??{$re}))*\]};
        for (;;) {
                last unless $tempstr =~ s/(\[\w+?\s*=\s*($re|\n|[^\[\]
+])+\])/assign($1)/gie;
        }
[download]

It's unfortunate that you modify things without knowing why they were done in the first place.

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Re^4: How to find the outermost pair of square brackets with a regex? by lokiloki (Beadle) on Jan 17, 2007 at 20:35 UTC
thanks... i guess part of the problem, and perhaps why i was using the s modifier, was because how can something like the following also match for newline? `[^\[\]]` since this matches everything BUT `[` or `]`, shouldn't it also match newline? if not, what modifier can force it to match such? i couldnt figure that out, which is why i opted for this alternation: `($re\|\n\|[^\[\]])+`	[reply] [d/l] [select]
Re^5: How to find the outermost pair of square brackets with a regex? by ikegami (Patriarch) on Jan 17, 2007 at 20:43 UTC
`[^\[\]]` does match newline. `>perl -le "print( qq{\n} =~ /[^\[\]]/ ?1:0) 1` [download]	[reply] [d/l] [select]