Am 27.01.2011 03:51, schrieb Wayne Young:
Hi Andy,
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<[email protected]> wrote:
Am 26.01.2011 07:34, schrieb Wayne Young:
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?
What about '\\\', the given substitute will not turn it into '\\\\'.
With spaces to make it readable:
First command:
:%s/ \\ \@<! \%( \\ \\ \) * \\ \\ \@! / \\ & / g
works as follows.
Find an odd number of backslashes (one or more):
\%( \\ \\ \) * \\
The above will find '\\\' at the start of '\\\\' (which is bad); make
sure no '\' follows the match:
\%( \\ \\ \) * \\ \\ \@!
Now the above will find '\\\' at the end of '\\\\' (after backtracking,
the next try starts one position to the right); make sure no '\'
precedes the match:
\\ \@<! \%( \\ \\ \) * \\ \\ \@!
Eventually, '\','\\\',... are found and '\\','\\\\',... are skipped.
The replace part is: a backslash plus the whole matched text.
Second command:
:%s/ \\ \@<! \%( \%( \\ \\ \) * \) \@> \\ / \\ & / g
This one makes use of \@> (rarely to be found in patterns).
Find an odd number of '\':
\%( \%( \\ \\ \) * \) \@> \\
The above will find '\\\', but not at the start of '\\\\', because
'\\\\' is matched by \%( \\ \\ \) * and the surrounding \@> disables
backtracking (which would allow a match for the following \\ ).
Now when the whole pattern (so far) fails, the next try starts one
position to the right, allowing a match for '\\\' at the end of '\\\\'.
Like with the first command, this is avoided with \\ \@<! .
Third command:
:%s/ \\ \\ \= / \\ \\ / g
This one processes each encountered backslash, no problem with
backtracking then.
Get help:
:h pattern
:h /\@!
:h sub-replace-special
etc.
--
Andy
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