(((a +(b& 255))& 255)与((a + b)& 255)相同吗? [英] Is ((a + (b & 255)) & 255) the same as ((a + b) & 255)?

问题描述

我正在浏览一些C ++代码，发现了这样的内容:

I was browsing some C++ code, and found something like this:

(a + (b & 255)) & 255

双重加倍让我很烦，所以我想到了:

The double AND annoyed me, so I thought of:

(a + b) & 255

(a和b是32位无符号整数)

(a and b are 32-bit unsigned integers)

我迅速编写了一个测试脚本(JS)来确认我的理论:

I quickly wrote a test script (JS) to confirm my theory:

for (var i = 0; i < 100; i++) {
    var a = Math.ceil(Math.random() * 0xFFFF),
        b = Math.ceil(Math.random() * 0xFFFF);

    var expr1 = (a + (b & 255)) & 255,
        expr2 = (a + b) & 255;

    if (expr1 != expr2) {
        console.log("Numbers " + a + " and " + b + " mismatch!");
        break;
    }
}

尽管脚本证实了我的假设(两个操作都相等)，但我仍然不相信它，因为1)随机和2)我不是数学家，我不知道我在做什么.

While the script confirmed my hypothesis (both operations are equal), I still don't trust it, because 1) random and 2) I'm not a mathematician, I have no idea what am I doing.

此外，对Lisp-y头衔感到抱歉.随时进行编辑.

Also, sorry for the Lisp-y title. Feel free to edit it.

推荐答案

它们是相同的.这是一个证明:

They are the same. Here's a proof:

首先记下身份(A + B) mod C = (A mod C + B mod C) mod C

让我们通过将a & 255表示为a % 256的代表来重述该问题.这是正确的，因为a是无符号的.

Let's restate the problem by regarding a & 255 as standing in for a % 256. This is true since a is unsigned.

所以(a + (b & 255)) & 255是(a + (b % 256)) % 256

这与(a % 256 + b % 256 % 256) % 256相同(我已经应用了上述标识:请注意，mod和%对于无符号类型是等效的.)

This is the same as (a % 256 + b % 256 % 256) % 256 (I've applied the identity stated above: note that mod and % are equivalent for unsigned types.)

这简化为(a % 256 + b % 256) % 256，它变为(a + b) % 256(重新应用身份).然后，您可以将按位运算符放回给

This simplifies to (a % 256 + b % 256) % 256 which becomes (a + b) % 256 (reapplying the identity). You can then put the bitwise operator back to give

(a + b) & 255

完成证明.

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