Re^2: Regex, capturing variables vs. speed

That's a great tool, and here's something I discovered by playing around. There's a difference between executing:

  perl -Mre=debug -le'"lahblahblahblah" =~ /(.?lah)$1{2}/'
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and

  perl -Mre=debug -le'"lahblahblahblah" =~ /(.?lah)\1{2}/'
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The first one matches on "lahblahblah", and the second one matches on "blahblahblah". Why do you suppose that is?

The debugging engine appears to see the $1 as an almost literal copy of the regexp in the brackets. Whereas the \1 seems to be looking for whatever was matched by the regexp in the brackets..

Update: Thanks to Aristotle for pointing out the error in the $1 interpretation.

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Re^3: Regex, capturing variables vs. speed by Aristotle (Chancellor) on Nov 02, 2005 at 04:39 UTC
The regex engine is not seeing any `$1`. A `$1` in a regex does not mean anything special, it’s just a variable that gets interpolated as a literal string. In your case, `$1` is empty when that pattern is compiled, so `/(.?lah)$1{2}/` becomes simply `/(.?lah){2}/`. You can see this if you read the compiler output carefully – the `CURLYX[0] {2,2}` applies to the parenthesised expression. Makeshifts last the longest.	[reply]

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Re^3: Regex, capturing variables vs. speed
by Aristotle (Chancellor) on Nov 02, 2005 at 04:39 UTC

The regex engine is not seeing any $1. A $1 in a regex does not mean anything special, it’s just a variable that gets interpolated as a literal string. In your case, $1 is empty when that pattern is compiled, so /(.?lah)$1{2}/ becomes simply /(.?lah){2}/. You can see this if you read the compiler output carefully – the CURLYX[0] {2,2} applies to the parenthesised expression.

Makeshifts last the longest.

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