swi-prolog 中的算术表达式 [英] Arithmetic expression in swi-prolog

问题描述

我有以下序言程序:

set_1(2).
p(X) :- set_1(X+1).

我正在使用适用于 i386 的 SWI-Prolog 5.10.4 版在此程序上运行查询 p(1).

I'm using SWI-Prolog version 5.10.4 for i386 to run the query p(1) on this program.

答案是错误".

我希望答案是真"，因为 set_1(X+1) 应该以 set_1(2) 为基础并用第一个事实解决.

I expect the answer to be 'true', because set_1(X+1) should be grounded as set_1(2) and resolved with the first fact.

为什么答案是错误的，我怎样才能得到真实"?

Why the answer is false and how can I get 'true' ?

推荐答案

如果您希望 X+1 在您的示例中与 2 统一，您需要使用 is 进行编码/2.

If you want X+1 to unify with 2 in your example, you'll need to code this using is/2.

本身 X+1 是一个有效的 Prolog 术语，但即使 X 与 1 统一，术语也变成了 1+1，而不是您期望的 2.

By itself X+1 is a valid Prolog term, but even when X is unified with 1, the term becomes 1+1, not the 2 you expected.

尝试:

p(X) :- Y is X+1, set_1(Y).

补充:可能值得指出的是，Prolog 在计算算术表达式方面的极端懒惰"允许我们将计算责任从 p/1 推到 set_1/1，代价是必须使谓词成为规则而不是简单的事实.

Added: It's probably worth pointing out that the extreme "laziness" of Prolog in evaluating arithmetic expressions allows us to push down responsibility for evaluation from p/1 into set_1/1, at the expense of having to make that predicate a rule rather than a simple fact.

1 ?- [user].
|: set_1(X) :- 2 is X.
|: p(X) :- set_1(X+1).
|: {Ctrl-D}
% user://1 compiled 0.00 sec, 3 clauses
true.

2 ?- p(1).
true.

谓词 is/2 并不是唯一强制算术表达式求值的 SWI-Prolog 内置函数.请参阅此处了解完整的概要.特别是谓词 =:=(使用中缀表示法)，比较两个表达式是否具有相同的评估，在某些情况下可能很有用.

Predicate is/2 is not the only SWI-Prolog built-in that compels arithmetic expression evaluation. See here for a complete rundown. In particular predicate =:= (with infix notation), comparing whether two expressions have equal evaluations, might be useful in some cases.

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