From: come <come.news@...>
Date: 2007-03-14T21:30:08+09:00
Subject: Re: ruby regex lookarounds?

Hi,

The syntax (?:re) isn't a lookbehind. It is a grouping form like (re)
but without capture. So (?:(re)) is the same as (re). Off course, you
could write something like (?:a(bc)), and you will get "bc" in $1, and
not "abc".

On 14 mar, 09:35, 7stud 7stud <dol...@excite.com> wrote:

> I see what's going on. (?:ab) is not a grouping: (?: and ) do not form a
> grouping that gets a $ variable.  If I write it as (?:(ab)), then ab is
> a grouping:
>
> irb(main):021:0> "abc" =~ /(?:(ab))(c)/
> => 0
> irb(main):022:0> $&
> => "abc"
> irb(main):023:0> $1
> => "ab"
> irb(main):024:0> $2
> => "c"
> irb(main):025:0>
>
> It still seems a little strange that $& contains the lookaround.
>
> --
> Posted viahttp://www.ruby-forum.com/.