From: dblack@...
Date: 2007-07-25T05:15:08+09:00
Subject: Re: Regex Question

Hi --

On Wed, 25 Jul 2007, seijin@gmail.com wrote:

> Oh, I see.  So using the "*" with just one ... errr... thing will
> never match it?  "banana" =~ /a*/ will always be 0 but "banana" =~ /
> a*n/ will be 1 (one) ?  It's just confusing because I think of
> "greedy" as it trying to match as many characters as possible.  So
> that although "*" tell it to find 0 or more that it would work hard to
> return the maximum number of matches.

Keep in mind, though, that it's looking for a pattern-match starting
at the left.  If it finds it, it's finished; it's not going to keep
scanning the string.  In the leftmost position, it finds a match for
a* -- namely, 0 occurrences of a.  So it's done.

Here's an example that might help clarify it:

   "xxyyxxxyyy" =~ /x*/   # 0

The longest stretch of x's is at position 5.  However, the job of the
match engine is not to find the maximum number of x's; it's to find
the first match for /x*/, starting at the left.  It finds that match
at position 0.  There are only two x's, but that's neither here nor
there.


David

-- 
* Books:
   RAILS ROUTING (new! http://www.awprofessional.com/title/0321509242)
   RUBY FOR RAILS (http://www.manning.com/black)
* Ruby/Rails training
     & consulting:  Ruby Power and Light, LLC (http://www.rubypal.com)