Re^2: How to find the outermost pair of square brackets with a regex?

good idea, but this won't work because there may be multiple pairs of outermost (i.e., same level) brackets. so, a greedy regular expression will grab beyond. i should have clarified this. for example:

$str = '
blah blah blah blah
blah
blah blah blah [blah
[blah blah] [blah blah blah
blah] blah]
blah blah
blah blah blah blah
blah
blah blah blah [blah
[blah blah] [blah blah blah
blah] blah]
blah blah
blah blah blah blah
blah
blah blah blah [blah
[blah blah] [blah blah blah
blah] blah]
blah blah
blah blah blah blah
blah
blah blah blah [blah
[blah blah] [blah blah blah
blah] blah]
blah blah';
[download]

in other words, there are multiple top-level bracket pairs, which, may or may not, contain additional pairs (etc).

btw, here is the code snippet that i finally used (which has some nuances that i didn't include in the original question), based on the previous suggestion:

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

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Re^3: How to find the outermost pair of square brackets with a regex? by ikegami (Patriarch) on Jan 17, 2007 at 16:25 UTC
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.	[reply] [d/l] [select]
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]
Re^3: How to find the outermost pair of square brackets with a regex? by lokiloki (Beadle) on Jan 17, 2007 at 07:23 UTC
I should point out that ikegami's suggestion by itself would surely have worked, but I couldn't get it to for the nuances that I needed, and so I added some possibly unnecessary cruft to account for my needs. I.e., I don't really know why `\|\n\|[^\[\]+])+\]` was necessary, but after many hours of trial and error, it was that which made everything (appear to) work.	[reply] [d/l]