对负数取模 [英] modulo operation on negative numbers

问题描述

模运算a%b返回a/b的余数，但对于负数则不这样做.

A modulo operation a%b returns the remainder for a/b but for negative numbers it does not do so.

#include <stdio.h>

int main(void) {
  int n=-4;
  printf("%d\n",n%3);
  return 0;
}

它应该返回2，因为3 *(-2)=-6小于-4且是3的倍数，但输出为-1. 为什么将(-a) mod b与-(a mod b)

It should return 2 as 3*(-2)=-6 is just smaller than -4 and a multiple of 3 but the output is -1. Why is it treating (-a) mod b same as -(a mod b)

推荐答案

作为一般规则，模和除法应满足方程式

As a general rule, the modulo and division should satisfy the equation

b * (a/b) + a%b == a

对于正数，很明显，这意味着a%b必须为正数.但是，如果a/b为负，则结果将四舍五入为零.

For positive numbers, it is obvious that this means that a%b must be a positive number. But if a/b is negative, then the result is rounded towards zero.

因此，以 a = -4， b = 3为例.我们知道 a / b = -1.3333，四舍五入为a/b == -1.从上面的方程式中，我们可以得到b * (-1) + a%b == a.如果我们插入a和b，则会得到-3 + a%b == -4，并且看到a%b必须为-1.

So take for instance a = -4, b = 3. We know that a/b = -1.3333, which rounded towards zero becomes a/b == -1. From the equation above, we have that b * (-1) + a%b == a. If we insert a and b, we get -3 + a%b == -4, and we see that a%b must be -1.

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