事实序言

Question

I would like to create a list of all the exam(A,B) contained in the knowledge base:

exam([a, 26]).
exam([b, 30]).
exam([c, 20]).
exam([d, 19]).

I know that i can simply use:

exams(L) :- findall(X, exam(X), L).

But since i want to write my own roles, i wrote the following:

exams([exam([A,B])]) :- exam([A,B]).
exams([exam([A,B])|Ex]) :- \+(member(exam([A,B]),Ex)), exams(Ex).

Why with a query like ?- exams(X). the system answers as follow:

?- exams(X).
X = [exam([a, 26])] ;
X = [exam([b, 30])] ;
X = [exam([c, 20])] ;
X = [exam([d, 19])] ;
false.

And doesn't give me any list with more than just one exam, also if the following query returns true:

?- exams([exam([a,26]),exam([b,30])]).
true .

Answer 1

a possibility is

exams(L) :- exams([], R), !, reverse(R, L).

exams(S, L) :- exam(E), \+ memberchk(exam(E), S), exams([exam(E)|S], L).
exams(L, L).

but, apart learning about Prolog control flow, I cannot see any value in such solution. Maybe you should elaborate the reason that force you to avoid findall/3...