s()谓词在Prolog中做什么? [英] What does the s() predicate do in Prolog?

问题描述

我一直在尝试学习Prolog，并且对谓词s()的功能完全感到困惑. 我看到它经常使用，并且关于Prolog的互联网资源很少，我找不到答案.

I have been trying to learn Prolog, and am totally stumped on what the predicate s() does. I see it used often and there is so little resources on the internet about Prolog that I cannot find an answer.

例如.

    /* sum(Is,S) is true if S is the sum of the list of integers Is.           */
    sum([],0).
    sum([0|Is],S):-sum(Is,S).
    sum([s(I)|Is], s(Z) ):-sum([I|Is],Z).

推荐答案

s/1本身并不执行任何操作，并且实际上不是谓词.它们只是术语，代表其论证的后继者.因此，s(0)用于表示0(即1)的后继者，s(s(0))用于表示s(0)(即2)的后继者，依此类推.它们在Prolog中如此广泛，因为Prolog是执行符号计算的一种相当不错的语言，而即使简单的算术运算也感到笨拙，这意味着它们并未与编程范例无缝集成.

s/1 does not do anything in itself, and it's not really a predicate. They are just terms, a representation of the successor of their argument. So, s(0) is used to represent the successor of 0 (i.e. 1), s(s(0)) is used to represent the successor of s(0) (i.e. 2), and so on and so forth. They are so widespread in Prolog because Prolog is quite fine a language to perform symbolic computation, whereas even simple arithmetic operations feel clunky, meaning that they are not seamlessly integrated with the programming paradigm.

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