in reply to Re: Using Look-ahead and Look-behind
in thread Using Look-ahead and Look-behind

I think you have found a bug in 5.10's regex handling. The lookahead's greediness or non-greediness should not matter, because it does not consume any characters. When used in a global match, patterns that do not consume characters should advance one character on each match. At least that's how I read the documentation.

The really interesting thing about your version is that you didn't make the capture non-greedy, you made it optional. You probably meant (.*?), which (in pre-5.10) will output empty strings every time. I haven't installed 5.10 myself, so I can't play with it right now.

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Re: Regex bug in 5.10 (was: Using Look-ahead and Look-behind)
by almut (Canon) on May 08, 2009 at 15:06 UTC
    ...which (in pre-5.10) will output empty strings every time.

    With 5.10.0, /(?=(.*?))/g; outputs one empty string.  And I can confirm the behavior reported by jds17 with /(?=(.*))/g.

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Re: Regex bug in 5.10 (was: Using Look-ahead and Look-behind)
by jds17 (Pilgrim) on May 08, 2009 at 19:47 UTC
    You are right, my change did not affect greediness. The bad thing is: now I don't understand why my proposed solution worked at all. Maybe someone can explain? I don't think the question is too important, but I like to use regular expressions and it bugs me a little if I cannot understand one (especially such a tiny one).

    I have read the documentation you have cited and it helped, so I played around some more and tried out the following, which only exchanges "+" for "*" in your original expression, really works as one would think and therefore would be my preferred solution, at least for Perl 5.10: 

    $_ = "Hello";
print "$1\n" while /(?=(.+))/g;

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    Output: 
    Hello
ello
llo
lo
o

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      Note that using + instead * is not equivalent. 

      $ perl -e'$_ = "Hello"; print "<$1>\n" while /(?=(.*))/g;'
<Hello>
<ello>
<llo>
<lo>
<o>
<>

$ perl -e'$_ = "Hello"; print "<$1>\n" while /(?=(.+))/g;'
<Hello>
<ello>
<llo>
<lo>
<o>

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