Question

这是我的prolog代码，它给了我以下错误：

错误：超出全局堆栈

hasA(S,A):- name(S,L1),
            compar(L1,A).
compar([H1,H2|T1],T1):- H1 == 1575, H2 == 1740.

neda(H,A):- hasA(H,Z1),
            append(A,[1575,1740],A),
            state(Z1,Z2),
            append(A,Z2,A).
state([H|T],T):- H == 32.
state(A,A).

有什么问题？你能帮帮我吗？

Answer 1

假设您在neda(H,A)为自由变量的情况下致电A，您最终会调用append(A,[1575,1740],A)并A为自由变量。但是，传统上，append实现为

append([],Ys,Ys).
append([X|Xs],Ys,[X|Zs]) :- append(Xs,Ys,Zs).

append(A,[1575,1740],A)与append([],Ys,Ys)的统一失败，因为一方面A需要[]，另一方面[1575,1740]需要append(A,[1575,1740],A)。因此，Prolog尝试将append([X|Xs],Ys,[X|Zs])与{A/[X|Xs], Ys/[1575,1740], Xs/Zs}统一，并以append(Xs,[1575,1740],Xs)成功，这导致调用[1575,1740]并重复该过程，即Prolog进入无限递归并最终运行没法了。

如果我猜测您的意图正确，首先希望将A追加到Z2和然后附加A1。这意味着您需要引入两个新变量，例如A2和A，以便在每个追加步骤后记录neda(H,A):- hasA(H,Z1), append(A,[1575,1740],A1), state(Z1,Z2), append(A1,Z2,A2).的状态：

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波斯语言处理序言

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