$string =~ s/(\d)-([A-Z])/$1$2/ig;    # my old way
$string =~ s/(?<=\d)-(?=[A-Z])//ig;   # my new way

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Unless im missing something, those lines should both accomplish the exact same thing. Im really becoming a fan of the lookahead / lookbehind version, it seems like it should run faster than the $1$2 way, but i know virtually nothing of the underlying workings of the regex engine.

One way to find out which is faster is to use the Benchmark module that comes with the standard perl distribution. However, if execution speed really isn't that critical, it all boils down being clear to both Perl and the programmer. And if it does what you need, it's more important to be clear to the programmer as that is who will have to look at it and make sense of it in the future.

Update: quick benchmarks of my own show the look{ahead,behind} method to be faster and significantly so on large strings. I would imagine this is because of the excessive copying inherent in the $1$2 method.

duff

In reply to Re: lookahead / lookbehind vs other regex methods by duff
in thread lookahead / lookbehind vs other regex methods by shemp

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