From: Anthony Durity <anthony.durity@...>
Date: 2006-02-07T02:01:29+09:00
Subject: Re: Merging regular expressions

Yes!

This can be done with alternation (which is what RegExp.union does)
but I really really really want to know which sub-regexp matched - I
can (maybe) sort out the derivative problem of conflicting sub-regexps
when the time comes.

So, think of the first requirement as wanting to know which sub-regexp
in an array of regexps that have been unioned or alternated together
triggers a match - the object is to save time iterating over an array
of regexps.

hmmm,
    Anthony

(I think racc must implent this algorithm somewhere, I know not where)

On 2/6/06, Robert Klemme <bob.news@gmx.net> wrote:
> Anthony Durity wrote:
> > Thanks Robert, thanks Dave,
> >
> > I see what ye both mean. I was more thinking along the lines of a
> > situation like this. To avoid iterating over the array, I would like
> > to compress the regexps into a large regexp while still preserving the
> > identity of each regexp within the meta-regexp, something akin to what
> > happens when the scanner of a lexer is created.
> > So, imagine this trivial situation...
> >
> > 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!
>
> Didn't you originally state that you wanted to ensure that not two regexps
> match at the same time?  I mean this is something different.  What you now
> state can be done with an alternation - but you'll only see one of your
> sub regexps match, i.e. /(foo)|(bar)/.
>
> > See what I mean? Will I hack the Ruby source? Can this be done in
> > Ruby? Will I have to do it in C?
>
> The first question is: can this be done with regexps?  First of all I'd
> like to hear the exact requirements before we go on with this.  I have the
> feeling that they are not clear yet - or at least that they have not
> become clear to me.
>
> Cheers
>
>    robert
>
>
>