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
最简单的答案:
1111 + 1 =(1)0000。所以1111一定是-1。那么-1 + 1 = 0。
理解这些对我来说是完美的。
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'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
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
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。
