You seem to think backtracking cannot occur within (?>PAT). That's not the case at all.

The purpose of (?>PAT) is to prevent the regex engine from trying to get PAT to match something else once it has already matched something. In short, backtracking through (?>PAT) causes it to fail.

You want 

/^\d+ _\d+ (?> (?:_\d+)? ) \w+ $/x

[download]

which can also be written as 

/^\d+ _\d+ (?:_\d+)?+ \w+ $/x

[download]

For "1_0_1" =~ /^\d+ _\d+ (?> (?:_\d+)? ) \w+ $/x, everything is straightforward until /\w+/ fails to match. At that point, the regex engine starts to backtrack.

  1. Backtracking through causes it (?>...) fail (as always).
  2. Backtracking through causes it \d+ fail (since it previously only matched only one digit).
  3. Backtracking through causes it _ fail (as always).
  4. Backtracking through causes it \d+ fail (since it previously only matched only one digit).
  5. Backtracking through causes it ^ fail (as always).
  6. The match fails.

In reply to Re: Unexpected regular expression match by ikegami
in thread Unexpected regular expression match by GrandFather

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