On Wed, Apr 27, 2005 at 08:46:53AM -0400, Joshua Gatcomb wrote:
> The problem is that in the regex version I use capturing parens to
> identify the character matched.  For the purposes of the problem I
> don't need to rely on the first character matched I just need to know
> 1.
> Without doing a lot of research into junctions, I thought the
> following would work:
> my $matches = any( @x_chars ) eq any( @y_chars );
> my $match = $matches.pick;

Perhaps the easiest way to explain the difficulty here is to note that
executing a relational op (i.e. returning a boolean) value on a junction
argument returns a junction of boolean values.  So, the C<eq> expression
above doesn't contain the intersection of chars that match, it's just
a junction of the (boolean) values returned by &infix:<eq> on each
pair of arguments from @x_chars and @y_chars.  Actually, since
there's two junction arguments, the result will be an C<any> junction
of C<any> junctions.

> The worst that could happen is that I find out    
> there isn't a way to get a what matched from an any() eq any()       
> comparison.

Not using the standard &infix:<eq>, no.  But I suppose one could get
close to what you're wanting with something like:

   sub &infix:<myeq>(Str $a, Str $b) { return ($a eq $b) ? $a : ''; }

and then

   my $matches = any( @x_chars ) myeq any( @y_chars );
   my $match = $matches.pick;

although since $match is any( any( ... ), any( ... ), any( ... ) )
I'm not sure if C<.pick> would end up "picking" a junction as opposed
to one of the inner values, or if it could "pick" one of the empty
string values.

Anyway, hope this helps clear up why the original wasn't working.


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