Re: regexp: different match, re=1 vs re=0

William Fugh wrote:

> On Thu, Aug 22, 2013 at 10:14 PM, Bram Moolenaar <Bram@moolenaar.net> wrote:
>
> >
> > William Fugh wrote:
> >
> > > Version: 7.4, gVim, MS-DOS/MS-Windows
> > > =========================================================
> > >
> > :%s#\[\(com\|ref\|ex\|p\|i\|u\|s\|b\)\]\zs\s\+\|\s\+\ze\[/\(com\|ref\|ex\|p\|i\|u\|s\|b\)\]\|[({]\zs\s\+\|\s\+\ze[)}]\|\s\zs\s\+\|\s\+\ze[,;.]\([^.,;]\)##g
> > >
> > ----------------------------------------------------------------------------------------------------------------
> > > the command line as above, one line for test as follows, <Space> means a
> > > space.
> > >
> > -----------------------------------------------------------------------------------------------------------------
> > > This<Space>is<Space><Space><Space>[a<Space>test.
> > > =========================================================
> > > Result of re=1: This<Space>is<Space>[a<Space>test.
> > > Result of re=0: This<Space>is<Space><Space>test.
> > > ==============================================================*
> > > Both of these regexp engines have the same result if the following
> > command
> > > line is used:-)*
> > >
> > ---------------------------------------------------------------------------------------------------------------------------
> > >
> > :%s#\[\(com\|ref\|ex\|p\|i\|u\|s\|b\)\]\zs\s\+\ze\|\zs\s\+\ze\[/\(com\|ref\|ex\|p\|i\|u\|s\|b\)\]\|[({]\zs\s\+\ze\|\zs\s\+\ze[)}]\|\s\zs\s\+\ze\|\zs\s\+\ze[,;.]\([^.,;]\)##g
> > >
> > ---------------------------------------------------------------------------------------------------------------------------
> >
> > That's a long regexp. Can you reproduce it with a (much) shorter one?
> >
>
> :%s#\s\+\ze\[/\|\s\zs\s\+##g "result: (re=1) <> (re=0
> :%s#\s\+\ze\[/\|\s\zs\s\+\ze##g "result: (re=1) == (re=0

Thanks, that makes the puzzle much simpler.
I suspect it has something to do with \ze in an alternative.

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