On Fri, 16 Apr 2010, Tony Mechelynck wrote:
> On 16/04/10 01:00, Benjamin R. Haskell wrote:
> > I encountered an oddity when trying to use the substitute() function
> > to prepend '# ' to the start of every line returned from a command.
> >
> > A very simplified example:
> >
> > I might expect
> > :echo substitute("a\na",'^','b','g')
> > to echo "ba\na"
> >
> > But, why does
> > :echo substitute("a\na",'\_^','b','g')
> > also echo "ba\na" -- instead of "ba\nba"?
> >
> > In the end, my ugly workaround was to use:
> > join(map(split(system(...),"\n"),'substitute(v:val,x,y,z)'),"\n")
> > where I wanted just:
> > substitute(system(...),x,y,z)
> >
>
> \_^ matches start-of-line. However, unlike the range of the
> :s[ubstitute] command, the first argument of the substitute() function
> is regarded by Vim as one line. It has therefore only one
> start-of-line -- at the very beginning of the string.
>
> What you can do is use '^\|\n\zs' as second argument of substitute(),
> to match the start of the first argument or (zero-length) whatever
> follows an embedded linefeed.
Clever. I like it.
In my case, I realized after I'd sent it that I didn't really need the
power of substitute() after I used the join-map-split version:
join(map(split(system(...),"\n"),'"# ".v:val'),"\n")
But, I prefer your workaround:
substitute(system(...),'^\|\n\zs','# ','g')
--
Thanks,
Ben
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