Re: Dynamic regex assertions, capturing groups, and parsers: joy and terror

I find this quite confusing as well.

Update: No this makes perfect sense. Thanks to dio for straightening me out.

$rx = qr{ (.) (??{ print $1 }) }x;
print "!" if "japhy" =~ $rx;
__END__
japhy
[download]

How does $1 end up being 'japhy' with this re? Interestingly, changing it to

$rx = qr{ (.) (??{ print $1; '' }) }x;
print "!" if "japhy" =~ $rx;
__END__
j!
[download]

makes things work out properly.

---
$world=~s/war/peace/g

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Re^2: Dynamic regex assertions, capturing groups, and parsers: joy and terror by diotalevi (Canon) on Oct 03, 2005 at 23:09 UTC
The result of (??{ print $1 }) is 1 because print() succeeded in writing to STDOUT. The regex that was then compiled by (??{ ...}) was "1" which then failed. So the (.) advanced over every character and printed them individually. The proper thing to do here would have been (?{ ... }) which will not affect regex matching.	[reply]
Re^3: Dynamic regex assertions, capturing groups, and parsers: joy and terror by demerphq (Chancellor) on Oct 04, 2005 at 09:13 UTC
Doh. Of course. I knew that the print returning 1 failed the match, but i didn't put two and two together to realize that was why all of the chars were printed. And I've used this technique deliberately before too. /gah. Thanks for the clue-by-four. :-) --- $world=~s/war/peace/g	[reply]