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 '\\\\'. > > 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 > > > -- > You received this message from the "vim_use" maillist. > Do not top-post! Type your reply below the text you are replying to. > For more information, visit http://www.vim.org/maillist.php > -- You received this message from the "vim_use" maillist. Do not top-post! Type your reply below the text you are replying to. For more information, visit http://www.vim.org/maillist.php
