Re: fund with findall

On 1/04/2014, at 10:03 AM, Ross Boylan wrote:

> Given
> foo(1, 2).
> foo(3, 2).
> foo(1, 10).
> 
> funky(Src, Dest, Matches) :-
> 	findall(x(Src, Dest), foo(Src, Dest), Matches).
> 
> Then funky(1, B, M) produces
> 
> M = [x(1, 2), x(1, 10)].
> 
> I find this slightly disturbing because it means that Dest got bound to
> 2 separate values, while remaining unbound at the end.

Well, yes.  By putting Dest in the template, you *asked*
for it to be existentially quantified.

Roughly,

   findall(Template, Generator, List) :-
      push a marker on a stack,
      (   call(Generator),
          push a copy of Template on the stack
          fail ; true
      ),
      pop results off the stack until the marker is reached,
      putting them in List.