From: Eero Saynatkari <ruby-ml@...>
Date: 2006-02-07T04:29:09+09:00
Subject: Re: Merging regular expressions

On 2006.02.07 04:26, Eero Saynatkari wrote:
> 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

Bah.
   
   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