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Do you know where your variables are?

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Lets dissect this into smaller problems.


I tried to simplify the case to avoid misunderstandings

DB<32> p "hello" =~ s/o*$/O/gr; hellOO DB<33> $_="hello"; s/o*$/O/g; print # for older Perls hellOO DB<34>

Surprise: the o is replaced twice.

Explanation so far

You and Hauke already explained that

  • pos isn't changing after the first match b/c of the zero-width of $
  • the empty o* is matching again

(And I agree that the referenced perlre#Repeated-Patterns-Matching-a-Zero-length-Substring needs a rewrite)

DB<41> $_="hello"; say pos,"($1)" while m/(o*$)/g; # pos doesn't c +hange 5(o) 5() DB<42> p "hello" =~ s/x*$/O/gr; # empty match ( +no x) helloO


Now, why is it surprising?

I think your case is that $ in combination with the /m modifier should act differently. Correct?

  • Would this be consistent?
  • Are there already examples of zero-width assertions who does it that way?
  • Are there work-arounds to achieve what you want? (i.e. skipping zero-length matches)

Here a guess for the last question

DB<44> p "hello\nfoo" =~ s/o*\n/O/gmr; hellOfoo DB<45> p "hello\nfoo\n" =~ s/o*\n/O/gmr; # added \n at the end of + input hellOfO DB<46>


Question @all: Is the problem better understood now? :)

Cheers Rolf
(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery


added more code


added headlines for structuring

) because empty patterns are always matching


DB<59> p "12345" =~ s/x*/ /gmr; 1 2 3 4 5 DB<60>

In reply to Re: You won't believe what this regular expression does! by LanX
in thread You won't believe what this regular expression does! by salva

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