🚀 go-pugleaf

> how do i take a predicate name as an argument and use it as a basis for
> selecting an operation on the other arguments?
>	[ e.g.: question('+', X, Y, Answer) :- Answer is X + Y . ]
> i would like question to be general enough to take any operator and perform
> it on X and Y resulting in Answer.

The general solution to this question in IBM-Prolog is
(using its Edinburgh syntax):

        question(Pred, X, Y, Answer) :- Answer is Pred(X,Y) .

or even more compactly using its standard syntax:

        question(Pred, X, Y, ? Pred(X,Y)) .

However, this solution is not readily available in other
Prologs because:

- Functor names can't be variables.  It would seem that this
  could be overcome by using =../2:

        question(Pred, X, Y, Answer) :-
                F =.. [Pred, X, Y],
                Answer is F .

- But general terms aren't allowed with "is" (at least, this is
  the case with Quintus 2.x; I assume it's true of others); if	
  you try this, you will get a run-time error for is/2. 

So, the answer would seem to be "it can't be done" in many
popular Prologs; you'll just have to list out all the arithmetic
operators yourself (there aren't too many, fortunately).

Peter Ludemann     IBM Prolog Development (these are my opinions, not IBM's)