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Re: is (?: ... ) capturing

by ysth (Canon)
on Sep 07, 2009 at 07:35 UTC ( #793895=note: print w/replies, xml ) Need Help??


in reply to is (?: ... ) capturing

Capturing refers to data that is put into $1, $2, etc. or returned from a list context m//. You aren't using the result of a capture there, so capturing or not isn't relevant. Even though (?: ) doesn't capture, it does consume part of the string and usually becomes part of what s/// will replace. You can avoid that with a lookbehind instead: s/(?<=\s)\w/_/g. Or, in 5.10+, the \K feature: s/\s\K\w/_/g. See perlre for details.
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Re^2: is (?: ... ) capturing
by jeanluca (Deacon) on Sep 07, 2009 at 07:45 UTC
    thnx a lot!!
    I'm using 5.6, so if I had to do something like
    my $str = "abc def 123 456" ; $str =~ s/(?:\s+)\w/_/g ;
    I would have had to do it differently, correct ? cheers
        s/(\s+)\w/${1}_/g
        simplifies to
        s/(\s)\w/${1}_/g
        Now you can use a zero-width lookbehind to get the speed boost of avoiding captures.
        s/(?<=\s)\w/_/g
        Or in 5.10+:
        s/\s\K\w/_/g
        ok, thnx again!!

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