c - 为什么 C 和 Ruby 中负整数的模运算符 (%) 的行为不同？

Question

我在 here 中运行一些代码.我试过 -40 % 3。它给了我输出 2。当我在 C 中执行相同的操作时，我得到:

int i = (-40) % 3
printf("%d", i);

输出是

-1

这两种语言如何在内部执行模运算？

Answer 1

Wiki说:

Given two positive numbers, a (the dividend) and n (the divisor), a modulo n (abbreviated as a mod n) is the remainder of the Euclidean division of a by n.
.... When either a or n is negative, the naive definition breaks down and programming languages differ in how these values are defined.

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现在的问题是为什么-40 % 3是2在 Ruby 中，或者换句话说，背后的数学原理是什么？

让我们从Euclidean division开始其中指出:

Given two integers a and n, with n ≠ 0, there exist unique integers q and r such that a = n*q + r and 0 ≤ r < |n|, where |n| denotes the absolute value of n.

现在请注意商的两个定义:

1. Donald Knuth描述了下限除法，其中商由下限函数定义 q=floor(a/n)和其余的 r是:

此处商 ( q ) 始终向下舍入(即使它已经是负数)，余数 ( r ) 与除数的符号相同。

1. 一些实现将商定义为:

q = sgn(a)floor(|a| / n)哪里sgn是符号函数。

和余数 ( r ) 与股息 ( a ) 的符号相同。

现在一切都取决于q :

• If implementation goes with definition 1 and define q as floor(a/n) then the value of 40 % 3 is 1 and -40 % 3 is 2. Which here seems the case for Ruby.
• If implementation goes with definition 2 and define q as sgn(a)floor(|a| / n), then the value of 40 % 3 is 1 and -40 % 3 is -1. Which here seems the case for C and Java.