我在上一门计算机系统课程，在某种程度上，我一直在为“二的补充”而挣扎。我想了解它，但我所读到的一切并没有让我明白它的含义。我读过维基百科的文章和其他各种文章，包括我的课本。
什么是2的补数，我们如何使用它，它如何在强制转换(从有符号到无符号，反之亦然)、逐位操作和位移位操作等操作中影响数字?
当前回答
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).
其他回答
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。
2的补码对于查找二进制值非常有用，但是我想到了一个更简洁的方法来解决这样的问题(从未见过其他人发布它):
以二进制为例:1101(假设空格“1”是符号)等于-3。
使用2的补码，我们可以这样做…翻1101到0010…加上0001 + 0010 ===>得到0011。0011的正二进制= 3。因此1101 = -3!
我意识到:
而不是所有的翻转和加法，你可以只做一个基本的方法来解决正二进制(假设0101)是(23 * 0)+(22 * 1)+(21 * 0)+(20 * 1)= 5。
用否定句做同样的概念!(稍微扭曲一下)
以1101为例:
对于第一个数字，用-(23 * 1)= -8代替23 * 1 = 8。
然后像往常一样，做-8 + (22 * 1)+ (21 * 0)+ (20 * 1)= -3
问题是“什么是“2的补码”?”
对于那些想要从理论上理解它的人(以及我试图补充其他更实际的答案)，简单的答案是:2的补码是对偶系统中不需要额外字符(如+和-)的负整数的表示。
2的补语:当我们用一个数字的1的补语加一个额外的1时，我们将得到2的补语。例如:100101，它的1的补足是011010和2的补足是011010+1 = 011011(通过与1的补足相加) 本文以图解的方式对此进行了解释。
补一词来源于完备性。在十进制世界中，数字0到9提供了一个数字或数字符号的补集(完整集)来表示所有的十进制数。在二进制世界中，数字0和1提供了一个数字的补数来表示所有二进制数。事实上，符号0和1必须用来表示所有东西(文本、图像等)以及正(0)和负(1)。 在我们的世界里，数字左边的空白被认为是零:
35=035=000000035.
In a computer storage location there is no blank space. All bits (binary digits) must be either 0 or 1. To efficiently use memory numbers may be stored as 8 bit, 16 bit, 32 bit, 64 bit, 128 bit representations. When a number that is stored as an 8 bit number is transferred to a 16 bit location the sign and magnitude (absolute value) must remain the same. Both 1's complement and 2's complement representations facilitate this. As a noun: Both 1's complement and 2's complement are binary representations of signed quantities where the most significant bit (the one on the left) is the sign bit. 0 is for positive and 1 is for negative. 2s complement does not mean negative. It means a signed quantity. As in decimal the magnitude is represented as the positive quantity. The structure uses sign extension to preserve the quantity when promoting to a register [] with more bits:
[0101]=[00101]=[00000000000101]=5 (base 10)
[1011]=[11011]=[11111111111011]=-5(base 10)
用作动词: 2的补语表示否定。这并不意味着消极。意思是如果负数变成正数;如果是正的就是负的。大小是绝对值:
if a >= 0 then |a| = a
if a < 0 then |a| = -a = 2scomplement of a
此功能允许使用先求负后加的有效二进制减法。 A -b = A + (-b)
1的补数的官方方法是每一位数用1减去它的值。
1'scomp(0101) = 1010.
这与逐个翻转或反转每一位是一样的。结果是- 0，这是不受欢迎的，所以给te 1的补码加上1就解决了这个问题。 要求2s的补，先求1s的补，然后加1。
Example 1 Example 2
0101 --original number 1101
1's comp 1010 0010
add 1 0001 0001
2's comp 1011 --negated number 0011
在这些例子中，否定也适用于符号扩展数。
添加: 1110进位111110进位 0110与000110相同 1111年 111111年 Sum 0101 Sum 000101
减法:
1110 Carry 00000 Carry
0110 is the same as 00110
-0111 +11001
---------- ----------
sum 0101 sum 11111
请注意，当使用2的补码时，数字左侧的空白区域对于正数用0填充，而对于负数用1填充。进位总是被加上，必须是1或0。
干杯
