Prolog: “findall” for limited number of solutions

Question

Say I want to find sum of all solutions to a predicate, I just can use

findall(L, find(L), Sols),

and just sum members of Sols.

But

Answer 1

A portable solution in ISO Prolog:

:- dynamic(find_n_solution/1).
:- dynamic(find_n_counter/1).

find_n(N, Term, Goal, Solutions) :-
    (   set_find_n_counter(N),
        retractall(find_n_solution(_)),
        once((
            call(Goal),
            assertz(find_n_solution(Term)),
            dec_find_n_counter(M),
            M =:= 0
        )),
        fail
    ;   findall(Solution, retract(find_n_solution(Solution)), Solutions)
    ).

set_find_n_counter(N) :-
    retractall(find_n_counter(_)),
    assertz(find_n_counter(N)).

dec_find_n_counter(M) :-
    retract(find_n_counter(N)),
    M is N - 1,
    assertz(find_n_counter(M)).

Using the sample predicate find/1 @ChristianF answer:

| ?- find_n(10, X, find(X), L).
L = [0,1,2,3,4,5,6,7,8,9]
yes

Note that this solution will still return a list of solutions even if there are less than the required number of solutions. Some Prolog compilers, including B-Prolog and ECLiPSe provide non-logical global variables that can be used to implement the counter but that would make the solution non-portable.