I can understand the pattern ifys0325 gave now (thanks ifys0325!) , but yours really scared me... All I can know is the last one with a "\=".
The first one seems to be: find a "\\*", without a "\" ahead, and without a "\\" followed?
And the second one: no "\" leads, but followed by a "\", and it matches "\\*" ?
I think I may wrong to parse those two, but I am not able to understand them at all...
-- On Thu, Jan 27, 2011 at 5:03 AM, Andy Wokula <anwoku@yahoo.de> wrote:
Am 26.01.2011 07:34, schrieb Wayne Young:What about '\\\', the given substitute will not turn it into '\\\\'.
Thanks very much. The second works!
I am still wondering how it work. It seems to replace the "
\%(\\\)\@<!\\\$(\\\)\@! " part to "\\\\". Would you help to explain the
meaning of the replaced part?
Better alternatives (all three supposed to do the same):
:%s/\\\@<!\%(\\\\\)*\\\\\@!/\\&/g
:%s/\\\@<!\%(\%(\\\\\)*\)\@>\\/\\&/g
:%s/\\\\\=/\\\\/g
The last one is short, but will not report a useful number of changes.
--
Andy
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