按位异或两个数字会得到数字的和或差

当我对任意两个数字进行异或时，我得到的是它们的差值或总和的绝对值。 我在谷歌上搜索了很多，试图找到任何相关的公式。但对此没有明显的公式或陈述。

Example:

10 XOR 2 = 1010 XOR 10 = 1000(8)
 1 XOR 2 =   01 XOR 10 =   11(3)

所有数字都是如此吗？

正如 Dukelings 的回答和 CmdrMoozy 的评论所示，这并不总是正确的。从你的帖子来看，至少是真的有时。那么这里就稍微详细一点的分析一下。

The +-side

显然，如果（但不仅仅是如果）(x & y) == 0 then (x ^ y) == x + y， 因为
x + y = (x ^ y) + ((x & y) << 1)

That accounts for 3³² cases (for every bit position, there are 3 choices that result in a 0 after the AND) where (x ^ y) == (x + y).

然后还有一些情况(x & y) != 0。这些案例正是这样的案例
(x & y) == 0x80000000，因为最高位的进位是唯一不影响任何内容的进位。

That adds 3³¹ cases (for 31 bit positions there are 3 choices, for the highest bit there is only 1 choice).

The --side

对于减法，有一个鲜为人知的恒等式x - y == (x ^ y) - ((~x & y) << 1).

That's really not too different from addition, and the analysis almost the same. This time, if (but not only if) (~x & y) == 0 then (x ^ y) == x - y. That ~ doesn't change the number of cases: still 3³². Most of them are different cases than before, but not all (consider y = 0, then x can be anything).

There are again 3³¹ extra cases, this time from (~x & y) == 0x80000000.

双方

The + and -双方并不脱节。有时，x ^ y = x + y = x - y。这只有在以下任一情况下才会发生y = 0 or y = 0x80000000. If y = 0, x可以是任何东西，因为(x & 0) == 0 and (~x & 0) == 0对全部x. If y = 0x80000000, x又可以是任何东西，这次是因为x & 0x80000000 and ~x & 0x80000000两者都可以算出 0 或0x80000000，两者都很好。

That gives 2³³ cases where x ^ y = x + y = x - y.

It also gives (3³² + 3³¹) * 2 - 2³³ cases where x ^ y is x + y or x - y or both, which is 4941378580336984 or in base 16, 118e285afb5158, which is also the answer given by this site.

That's a lot of cases, but only roughly 0.02679% of the total space of 2⁶⁴.