cut in SWI-Prolog

I have two queries:

a) ?- G = (member(X, [a, b]), !, write(X)), call(G).

b) ?- G = (member(X, [a, b]), call(!), write(X)), call(G).

My question is: what is the difference between scope of "!" in (a) and
(b) in these queries (as in both of them G is metacalled) ?

In message <9abc3ca2.0412010438.7f2f4e94@posting.google.com>, Budyn
<budyn@zamek.gda.pl> writes
>I have two queries:
>
>a) ?- G = (member(X, [a, b]), !, write(X)), call(G).
>
>b) ?- G = (member(X, [a, b]), call(!), write(X)), call(G).
>
>My question is: what is the difference between scope of "!" in (a) and
>(b) in these queries (as in both of them G is metacalled) ?

You can see the same effect more simply with
?- member(X, [a, b]), !, write(X).
which just has one solution, and
?- member(X, [a, b]), call(!), write(X).
which is resatisfiable.

"!" cuts back to the call (or redo) of the predicate which contains it.
In
?- G = (member(X, [a, b]), !, write(X)), call(G).
this is call(G).  In
?- member(X, [a, b]), !, write(X).
it is the command-line query.  In
?- G = (member(X, [a, b]), call(!), write(X)), call(G).
and
?- member(X, [a, b]), call(!), write(X).
it is the call to 'call'.

Nick
--
Nick Wedd    nick@maproom.co.uk