Re: You won't believe what this regular expression does!

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]

Comment on Re: You won't believe what this regular expression does! Select or Download Code

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Re^2: You won't believe what this regular expression does! by haukex (Archbishop) on Feb 25, 2021 at 14:42 UTC
Are there work-arounds to achieve what you want? I sometimes use `(?:\n\|\z)` to be explicit that I want the line endings to be consumed by the engine.	[reply] [d/l]
Re^3: You won't believe what this regular expression does! by LanX (Saint) on Feb 25, 2021 at 15:15 UTC
Thanks! :) But please note the second `fOO` `DB<55> p "hello\nfoo" =~ s/o*(?:\n\|\z)/O/gmr; hellOfOO DB<56>` [download] I'm busy right now, but I seem to remember that one could use features for atomic matches... I'll try later...° Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery} update °) nah doesn't help, since it's not a backtracking problem.	[reply] [d/l] [select]
Re^4: You won't believe what this regular expression does! by haukex (Archbishop) on Feb 26, 2021 at 08:26 UTC
But please note the second fOO Yes, good point! I think the main question is what the intent of the regex is. If it's "replace any `o`'s at the end of each line", then the better solution is, as you said, `/o+$/`, and using `o*` is the "mistake".	[reply] [d/l] [select]
Re^5: You won't believe what this regular expression does! by salva (Canon) on Feb 26, 2021 at 11:55 UTC
Re^6: You won't believe what this regular expression does! by haukex (Archbishop) on Feb 26, 2021 at 13:16 UTC
Re^6: You won't believe what this regular expression does! by LanX (Saint) on Feb 26, 2021 at 13:17 UTC
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Do you know where your variables are?
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Re: You won't believe what this regular expression does!

Simplification

Explanation so far

Disappointments

Workarounds

Meta

edit

update

update