-2
下面是我用事实:如何摆脱在这种情况下,未分配的指针?
edge(route(fra), route(tog), walk).
edge(route(fra), route(105), walk).
edge(route(tog), route(togc), bus).
edge(route(togc), route(sub), walk).
edge(route(sub), route(ag), metro).
edge(route(ag), route(n), metro).
edge(route(n), route(a), metro).
edge(route(a), route(r), metro).
edge(route(re), route(m), metro).
edge(route(m), route(v), walk).
edge(route(105), route(t), bus).
edge(route(t), route(vi), metro).
edge(route(vi), route(m), metro).
edge(route(t), route(3), walk).
edge(route(3), route(m), bus).
而且断言:
seek(route(v), _).
seek(route(R1), [[R2, How]|Res]) :-
edge(route(R1), route(R2), How),
seek(route(R2),Res).
,当我问查询?- seek(route(fra), X).
应该在列表中返回R2
和How
,但Res
在这种情况下,仍然未分配的......我不知道我怎样才能解决,为了使该查询只返回包含R2
列表的列表。任何帮助将不胜感激!