双重AND惹恼了我，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)

我很快写了一个测试脚本(JS)来证实我的理论:

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;
    }
}

虽然脚本证实了我的假设(两个操作相等)，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.

另外，对于 Lisp-y 标题感到抱歉.随意编辑它.

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

推荐答案

它们是一样的.这是一个证明:

They are the same. Here's a proof:

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

让我们通过 a & 来重申这个问题.255 代表 a % 256.这是真的，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:注意 mod 和 % 等价于无符号类型.)

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 变成 (a + b) % 256 (重新应用身份).然后你可以把按位运算符放回去给

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