在 Prolog 中更改值 [英] Change value in Prolog

问题描述

我是 Prolog 的新手，我想更改从列表中提取的变量的值.最初，变量是 n，然后在某些情况下我想将其更改为 a.但是使用 (is)/2 是行不通的，因为它只对数字进行操作.

I am new to Prolog, and I want to change the value of a variable, which is extracted from a list. Initially, the variable is n, then on some occasions I would like to change it to a. But using (is)/2 won't work because it only operates on numbers.

有没有简单的方法来做到这一点?假设我的代码如下所示:

Is there an easy way to do this? Suppose my code looks something like this:

change([H|T]) :- set H to a,change(T).
change([]).

注意 H 已经设置为 n，所以目标 H = a 失败，因为 n 和a不能统一.

Notice H has already been set to n, so the goal H = a fails because n and a cannot be unified.

推荐答案

您在学习 prolog 时遇到了一个关键问题，即它不像过程语言那样工作.

You're hitting the key issue when learning prolog that it doesn't work like procedural languages.

prolog 中的变量是一个变量，它可以具有任何值，但在计算中的任何时候，如果变量已经统一，那么除非 prolog 回溯，否则它不能改变.

A variable in prolog is a variable in the sense that it can have any value, but at any point in a computation if the variable has been unified then it cannot change unless prolog backtracks.

所以，你不能简单地取一个列表，例如 [m, n, o, p] 并将其更改为 [m, a, o, p].您必须构建一个新列表.

So, you cannot simply take a list, such as [m, n, o, p] and change it to be [m, a, o, p]. You have to construct a new list.

方法如下:

replace_n_with_a([], []).
replace_n_with_a([n|X], [a|Y]) :- replace_n_with_a(X, Y).
replace_n_with_a([H|X], [H|Y]) :- H = n, replace_n_with_a(X, Y).

这三个谓词获取一个列表并构建一个新列表，但只要找到它就将 n 替换为 a.原始列表没有改变，但我现在有了一个新列表，可以将其传递给我的代码的下一部分.

These three predicates take a list and build a new one, but swap n for a whenever it finds it. The original list hasn't changed, but I now have a new one that I can pass to the next part of my code.

要运行上面的代码，你可能有这个:

To run the above code you may have this:

?- replace_n_with_a([m, n, o, p], Xs), write(Xs), nl.

我得到这个结果:

[m, a, o, p]
Yes.

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