我在Reddit上读到jng的一篇精彩的解释，用里程表做类比。

It is a useful convention. The same circuits and logic operations that add / subtract positive numbers in binary still work on both positive and negative numbers if using the convention, that's why it's so useful and omnipresent. Imagine the odometer of a car, it rolls around at (say) 99999. If you increment 00000 you get 00001. If you decrement 00000, you get 99999 (due to the roll-around). If you add one back to 99999 it goes back to 00000. So it's useful to decide that 99999 represents -1. Likewise, it is very useful to decide that 99998 represents -2, and so on. You have to stop somewhere, and also by convention, the top half of the numbers are deemed to be negative (50000-99999), and the bottom half positive just stand for themselves (00000-49999). As a result, the top digit being 5-9 means the represented number is negative, and it being 0-4 means the represented is positive - exactly the same as the top bit representing sign in a two's complement binary number. Understanding this was hard for me too. Once I got it and went back to re-read the books articles and explanations (there was no internet back then), it turned out a lot of those describing it didn't really understand it. I did write a book teaching assembly language after that (which did sell quite well for 10 years).

2对给定数的补数是1与1的补数相加得到的数。

假设我们有一个二进制数:10111001101

它的1的补位是:01000110010

它的2的补数是:01000110011

2's complement is essentially a way of coming up with the additive inverse of a binary number. Ask yourself this: Given a number in binary form (present at a fixed length memory location), what bit pattern, when added to the original number (at the fixed length memory location), would make the result all zeros ? (at the same fixed length memory location). If we could come up with this bit pattern then that bit pattern would be the -ve representation (additive inverse) of the original number; as by definition adding a number to its additive inverse always results in zero. Example: take 5 which is 101 present inside a single 8 bit byte. Now the task is to come up with a bit pattern which when added to the given bit pattern (00000101) would result in all zeros at the memory location which is used to hold this 5 i.e. all 8 bits of the byte should be zero. To do that, start from the right most bit of 101 and for each individual bit, again ask the same question: What bit should I add to the current bit to make the result zero ? continue doing that taking in account the usual carry over. After we are done with the 3 right most places (the digits that define the original number without regard to the leading zeros) the last carry goes in the bit pattern of the additive inverse. Furthermore, since we are holding in the original number in a single 8 bit byte, all other leading bits in the additive inverse should also be 1's so that (and this is important) when the computer adds "the number" (represented using the 8 bit pattern) and its additive inverse using "that" storage type (a byte) the result in that byte would be all zeros.

 1 1 1
 ----------
   1 0 1
 1 0 1 1 ---> additive inverse
  ---------
   0 0 0

简单来说，2的补码是一种在计算机内存中存储负数的方法。而正数则存储为普通二进制数。

让我们考虑这个例子，

计算机使用二进制数字系统来表示任何数字。

x = 5;

这表示为0101。

x = -5;

当计算机遇到-号时，它会计算出它的2的补数并存储它。

也就是说，5 = 0101，它的2的补是1011。

计算机处理数字的重要规则是，

如果第一位是1，那么它一定是负数。 如果除第1位之外的所有位都是0，那么它就是一个正数，因为在数字系统中没有-0(1000不是-0，而是正8)。 如果所有的位都是0，那么它就是0。 否则就是正数。

Two的补码是一种存储整数的聪明方法，因此常见的数学问题很容易实现。

为了理解，你必须把数字想象成二进制。

它基本上是说，

对于0，用所有的0。 对于正整数，开始计数，最大值为2(位数-1)-1。 对于负整数，做完全相同的事情，但是切换0和1的角色并开始倒数(所以不是从0000开始，而是从1111开始——这是“补”部分)。

让我们尝试一个4位的迷你字节(我们称之为1/2个字节)。

0000 -零 0001 - 1 0010 - 2 0011 - 3 0100到0111,4点到7点

这是我们目前能找到的阳性结果。23-1 = 7。

负面影响:

1111 - 1 1110 - 2 1101 - 3 1100到1000 - - 4到- 8

注意，负数(1000 = -8)有一个额外的值，而正数没有。这是因为0000用于表示零。这可以看作是计算机的数轴。

区分正数和负数

这样一来，第一个位就扮演了“符号”位的角色，因为它可以用来区分非负的十进制值和负的十进制值。如果最高有效位是1，那么二进制就可以说是负的，如果最高有效位(最左边)是0，就可以说十进制值是非负的。

“符号量级”的负数只是将它们的正数对应的符号位颠倒了，但这种方法必须处理将1000(一个1后面跟着所有的0)解释为“负零”，这是令人困惑的。

“1的补”负数只是它们的正数的位补，这也导致了“负零”和1111(都是1)的混淆。

除非你的工作非常接近硬件，否则你可能不需要处理个位补或符号幅度整数表示。

让我们用8位的二进制形式得到答案10 - 12: 我们要做的是10 + (-12)

我们需要用12的恭维部分减去10。 12的二进制值是00001100。 10的二进制值是00001010。

为了得到12的赞美部分，我们只需要把所有的位反转，然后加1。 12的二进制反转是11110011。这也是逆码(一个人的补码)。 现在我们需要加一个，现在是11110100。

所以11110100是12的赞美!这样想很简单。

现在你可以用二进制形式来解决上面的10 - 12问题了。

00001010
11110100
-----------------
11111110