From: Eero Saynatkari <ruby-ml@...>
Date: 2006-02-07T04:26:13+09:00
Subject: Re: Merging regular expressions

On 2006.02.07 02:24, Anthony Durity wrote:
> Jacob, Brilliant!
> 
> But... see my last post, i want to avoid iteration by merging all
> regexps into one large regexp. However, this is absolutely beautiful
> and correct. Taken with the replies I got earlier this is a great
> start.

If at all possible, I would use Strings:

  re      = ['foo', 'bar', 'baz', 'quux']
  regexp  = re.map {|r| "(#{r})"}.join '|'
  result  = 'baz baa foo guggeli quux'.scan regexp

  p result

You may need to tweak the regexp/builder a bit to
for example ensure there is whitespace on either 
side or something.  

> 
> Question... What would the following code output?
>
> r = []
> r << /aaa/
> r << /[ab][ab][ab]/
> m = Regexp.union( *r )
> m =~ "aaa" #  => ?
> 
> (I think I basically want to be able to twiddle with the finite
> automata the regexps make when they are compiled/created, I dunno if
> this is possible)
> 
> Later,
>     Anthony
> 
> On 2/6/06, Jacob Fugal <lukfugl@gmail.com> wrote:
> > On 2/6/06, Anthony Durity <anthony.durity@gmail.com> wrote:
> > > r[0] = /a/
> > > r[1] = /b/
> > > m = Regexp.union(r0.source, r1.source)
> > > # _but_ when I go to check s with
> > > m =~ s
> > > # if it matches, i want it to tell me which of the original r[i] would have matched!
> >
> > <skip union code />
> >
> > Jacob Fugal


E