Quoting randomizer <[EMAIL PROTECTED]>:
I am using a simple kb from a book "Learn prolog now!",
wizard(ron).
hasWand(harry).
quidditchPlayer(harry).
wizard(X) :- hasBroom(X),hasWand(X).
hasBroom(X) :- quidditchPlayer(X).
gnu prolog answers question wizard(harry). "no", and swi prolog answers
"yes".
anyone has an idea why does it happen?
--
Hi,
GnuProlog is very close to ISO prolog in most of its concepts. In ISO prolog,
the clauses of a predicate P/N must be contiguous!
In you example, wizard/1 is not. That's why the second clause was ignored.
| ?- consult('user').
compiling user for byte code...
wizard(ron).
hasWand(harry).
quidditchPlayer(harry).
wizard(X) :- hasBroom(X),hasWand(X).
user:4: warning: discontiguous predicate wizard/1 - clause ignored
hasBroom(X) :- quidditchPlayer(X).
user compiled, 6 lines read - 595 bytes written, 13840 ms
yes
... See the WARNING above.
The listing predicate gives the first clause only.
| ?- listing(wizard).
wizard(ron).
yes
So, you have to use the directive "contiguous" before the first clause of your
predicate :
| ?- consult('user').
compiling user for byte code...
:- discontiguous(wizard/1).
wizard(ron).
hasWand(harry).
quidditchPlayer(harry).
wizard(X) :- hasBroom(X),hasWand(X).
hasBroom(X) :- quidditchPlayer(X).
user compiled, 7 lines read - 801 bytes written, 34422 ms
yes
| ?- listing(wizard).
wizard(ron).
wizard(A) :-
hasBroom(A),
hasWand(A).
yes
| ?- wizard(harry).
yes
Good luck.
Ali.
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