序言和清单统一 [英] Prolog and List Unification

问题描述

我正在努力加深我对Prolog的理解，以及它如何处理统一性.在这种情况下，它如何处理与列表的统一.

I'm trying to further my understanding of Prolog, and how it handles unification. In this case, how it handles unification with lists.

这是我的知识库；

member(X, [X|_]).
member(X, [_|T]):- member(X, T).

如果我正确地理解了该过程.如果member(X, [X|_])不是true，则进入递归规则，如果X在列表T中，则[_|T]与T统一.

If I'm understanding the process correctly. If member(X, [X|_]) is not true, then it moves into the recursive rule, and if X is in list T, then [_|T] is unified with T.

那么我的递归谓词中的匿名变量会怎样?它会被丢弃吗?我很难理解列表的确切统一过程，因为[_|T]是两个变量，而不是一个.我只是想弄清楚统一过程如何与列表一起精确地工作.

So what happens to the anonymous variable in my recursive predicate? Does it get discarded? I'm having difficulty understanding the exact unification process with lists, as [_|T] is two variables, rather than one. I'm just trying to figure out how the unification process works precisely with lists.

推荐答案

假定_是Y

member(X, [Y|T]):- member(X, T).

这就是True，而不管Y.现在，您正在返回" member(X, T).换句话说，您将丢弃Y并返回" member(X, T).

Then this is True regardless Y. Now you are "returning" member(X, T). In other words, you are discarding Y and "returning" member(X, T).

_意味着，无论它是什么，都将忽略该变量.

_ means, whatever it is, ignore that variable.

_就像其他任何变量一样，除了您看到的每个变量都是 视为不同的变量，Prolog不会向您显示它的含义 与统一.那里没有特殊的行为.如果让您感到困惑 关于行为，只需发明一个全新的变量并将其放入 在那里看看它做什么.

The _ is just like any other variable, except that each one you see is treated as a different variable and Prolog won't show you what it unifies with. There's no special behavior there; if it confuses you about the behavior, just invent a completely new variable and put it in there to see what it does.

在这种情况下，您的函数将检查列表中是否存在给定元素，因此，获取列表中的第一个元素，检查是否相等，如果不相等，则丢弃该元素并继续前进.

In your case, your function check if a given element exists on a list, so, you take first element of the list, check if is equal, if not, you discard that element and moves on.

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