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http://bugs.exim.org/show_bug.cgi?id=1315
--- Comment #8 from Philip Hazel <ph10@???> 2012-11-09 16:10:06 ---
On Fri, 9 Nov 2012, Zoltan Herczeg wrote:
> > 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)
Ooops! Yes, indeed I did. Thinko. Sorry about that.
Philip
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