我喜欢lavinio的回答，但变换部分增加了一些复杂性。通常情况下，可以选择在保留符号位的情况下移动位，或者不保留符号位。这是将数字处理为有符号数字(-8到7表示小块，-128到127表示字节)或全范围无符号数字(0到15表示小块，0到255表示字节)之间的选择。

我喜欢lavinio的回答，但变换部分增加了一些复杂性。通常情况下，可以选择在保留符号位的情况下移动位，或者不保留符号位。这是将数字处理为有符号数字(-8到7表示小块，-128到127表示字节)或全范围无符号数字(0到15表示小块，0到255表示字节)之间的选择。

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).

Two的补语主要用于以下原因:

避免0的多个表示形式 避免在溢出的情况下跟踪进位(如补位)。 进行简单的加法和减法运算变得很容易。

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

我想知道是否有比维基百科上的文章更好的解释。

你试图用2的补表示法解决的基本问题是存储负整数的问题。

首先，考虑一个存储在4位的无符号整数。您可以拥有以下内容

0000 = 0
0001 = 1
0010 = 2
...
1111 = 15

这些是无符号的，因为没有指示它们是负的还是正的。

符号大小和多余符号

要存储负数，您可以尝试一些方法。首先，您可以使用符号幅度表示法，它将第一个位指定为符号位来表示+/-，其余位表示幅度。还是用4位假设1代表- 0代表+那么你就有

0000 = +0
0001 = +1
0010 = +2
...
1000 = -0
1001 = -1
1111 = -7

所以，你看到问题了吗?我们有正0和负0。更大的问题是二进制数的加减法。使用符号幅度进行加减法的电路将非常复杂。

是什么

0010
1001 +
----

?

另一个系统是过量符号。你可以存储负数，你可以摆脱两个0的问题但加减法仍然很困难。

于是就有了二的补。现在您可以存储正整数和负整数，并相对轻松地执行算术。有许多方法可以将一个数转换为二的补数。这是一个。

将十进制转换为二的补数

将数字转换为二进制(暂时忽略符号) 例如，5是0101，-5是0101 如果这个数字是正数，那么你就完成了。 例5是二进制的0101，使用二的补符号。 如果数字是负的，那么 3.1求补(0和1的倒数) 例如，-5是0101，所以找到补语是1010 3.2补数1010 + 1 = 1011加1 因此，2的补数-5等于1011。

那么，如果你想用二进制写2 +(-3)呢?2 +(-3) = -1。 如果你用符号的大小来加这些数，你需要做什么?0010 + 1101 = ?

使用2的补码，想想会有多简单。

 2  =  0010
 -3 =  1101 +
 -------------
 -1 =  1111

将2的补数转换为十进制

将1111转换为十进制:

这个数从1开始，所以它是负的，所以我们找到1111的补数，也就是0000。 0000加上1，得到0001。 将0001转换为十进制，即1。 应用符号= -1。

哒哒！