Re^2: A regexp to parse nested brackets containing strings

So you can't make $InnerRe anchorred at the beginning, and not anchorred at the beginning, all in the same match invocation.

It's not a problem. My solution even does this.

$\ = "\n";

my $re;
$re = qr/ a (??{ $re }) c | b /x;

print 'aabcc'  =~ /^$re/ ||0;  # 1
print '!aabcc' =~ /^$re/ ||0;  # 0  First call is anchored to start.
print 'a!abcc' =~ /^$re/ ||0;  # 0  Recursive call is anchored to pos.
print 'aa!bcc' =~ /^$re/ ||0;  # 0  Recursive call is anchored to pos.
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Re^3: A regexp to parse nested brackets containing strings by rodion (Chaplain) on Sep 24, 2006 at 08:27 UTC
Looks like I wasn't specific enough in my writeup. Yes, you can put a "^" anchor in the top-level invocation of $re, but not in the $re itself. For the OP's original problem, where he is looking for the innermost parenthesese that are not inside a quote, you want the "^L in the $re. If you don't put it there, the invocation of $re from within the first invocation of $re can just skip over text to find the parentheses within the quotes. However, "^" doesn't do what you need there, because it doesn't mean the beginning of where $re was invoked in each recursive iteration, it means the beginning of this invocation of the regex parser. At least that's what it looks like from the behavior.	[reply]
Re^4: A regexp to parse nested brackets containing strings by ikegami (Patriarch) on Sep 24, 2006 at 12:25 UTC
If you don't put it there, the invocation of $re from within the first invocation of $re can just skip over text to find the parentheses within the quotes. No, it can't skip. That's what I was showing. If it could skip, all the regexp in my parent post would match, since they would skip over the `!`. When using `(??{ $re })`, matching $re is anchored at the current pos..	[reply] [d/l] [select]