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http://bugs.exim.org/show_bug.cgi?id=1315
--- Comment #7 from Zoltan Herczeg <hzmester@???> 2012-11-09 13:18:50 ---
> This pattern does that: \r(?<!\n)
I think Philip wanted a forward assertion, not a backward, since this pattern
search for a \r character, and moves the position by one character ahead, and
checks that the last character is not \n, which is always true (\r != \n).
The correct form of this patern is: \r(?!\n)
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