comment on

Lets dissect this into smaller problems.

Simplification

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>
[download]

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
[download]

Disappointments

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)

Workarounds

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>
[download]

edit

added more code

update

added headlines for structuring

°) because empty patterns are always matching

compare

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

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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