负数的mod正在融化我的大脑 [英] Mod of negative number is melting my brain

问题描述

我试图mod的一个整数得到一个数组的位置，这样它会循环圆。做 I％
arrayLength 工作正常，正数，但对于负数这一切都错了。

I'm trying to mod an integer to get an array position so that it will loop round. Doing i % arrayLength works fine for positive numbers but for negative numbers it all goes wrong.

 4 % 3 == 1
 3 % 3 == 0
 2 % 3 == 2
 1 % 3 == 1
 0 % 3 == 0
-1 % 3 == -1
-2 % 3 == -2
-3 % 3 == 0
-4 % 3 == -1

所以我需要的实现。

so i need an implementation of

int GetArrayIndex(int i, int arrayLength)

这样

GetArrayIndex( 4, 3) == 1
GetArrayIndex( 3, 3) == 0
GetArrayIndex( 2, 3) == 2
GetArrayIndex( 1, 3) == 1
GetArrayIndex( 0, 3) == 0
GetArrayIndex(-1, 3) == 2
GetArrayIndex(-2, 3) == 1
GetArrayIndex(-3, 3) == 0
GetArrayIndex(-4, 3) == 2

我以前做了这一点，但由于某种原因，今天是融化我的大脑:(

I've done this before but for some reason it's melting my brain today :(

推荐答案

我总是用我自己的 MOD 函数，定义为

I always use my own mod function, defined as

int mod(int x, int m) {
    return (x%m + m)%m;
}

当然，如果你不屑大鱼大肉的两个的调用模操作，你可以把它写成

Of course, if you're bothered about having two calls to the modulus operation, you could write it as

int mod(int x, int m) {
    int r = x%m;
    return r<0 ? r+m : r;
}

或其变体。

它工作的原因是，×％m为始终在范围[-m + 1，M-1]。所以，如果在所有为负，加米它将把它的正数范围，而不改变其值模M

The reason it works is that "x%m" is always in the range [-m+1, m-1]. So if at all it is negative, adding m to it will put it in the positive range without changing its value modulo m.

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