Prolog中的Divison和其余部分 [英] Divison and remainder in Prolog

问题描述

试图找出如何写一个递归谓词divide_by（X，D，I，R），它将正整数X和除数D作为输入，并将答案作为整数I部分和其余部分返回然而，R，我似乎无法理解Prolog。我该怎么做呢？

Trying to figure out how to write a recursive predicate divide_by(X, D, I, R) that takes as input a positive integer X and a divisor D, and returns the answer as the whole number part I and the remainder part R, however, I can't seem to get my head around Prolog. How would I go about doing this?

推荐答案

这里有预定义的可评估仿函数。

There are predefined evaluable functors for this.

（div）/ 2 和（mod）/ 2 总是四舍五入。由LIA-1，Knuth等推荐。

(div)/2 and (mod)/2 always rounding down. Recommended by LIA-1, Knuth etc.

（//）/ 2 和 （rem）/ 2 向零舍入（实际上，它是实现定义的，但所有当前实现都是这样做的）。您可以通过 current_prolog_flag（integer_rounding_function，F）询问，这在当前实现中给出 towards_zero 。

(//)/2 and (rem)/2 rounding toward zero (actually, it's implementation defined, but all current implementations do it like that). You can ask this via current_prolog_flag(integer_rounding_function, F) which gives in current implementations toward_zero.

这些对之间的差异仅在涉及负数时显示。这是一种人们喜欢的宗教战争。 ISO / IEC 10967：2012语言无关算术（vl.LIA-1）仅提供舍入由于误用的倾向（C.5.1.2.2）of towards_zero，而Fortran和追随者喜欢C转向towards_zero，称之为algebraic（6.5.5）。另请参阅：
优势使用截断朝向负无穷大而不是零

The difference between those pairs shows only when negative numbers are involved. It is kind of a religious war which one to prefer. ISO/IEC 10967:2012 Language independent arithmetic (vl. LIA-1) only provides rounding down "due to proneness for erroneous use" (C.5.1.2.2) of toward_zero, whereas Fortran and followers like C go for toward_zero calling it "algebraic" (6.5.5). See also: Advantages of using truncation towards minus infinity vs towards zero

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