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

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。

其他回答

两人的补足(托马斯·芬利)

我把所有位的倒数加1。编程:

  // In C++11
  int _powers[] = {
      1,
      2,
      4,
      8,
      16,
      32,
      64,
      128
  };

  int value = 3;
  int n_bits = 4;
  int twos_complement = (value ^ ( _powers[n_bits]-1)) + 1;

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

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。

想象一下,你有有限数量的比特/比特/数字等等。将0定义为所有数字都为0,并自然向上计数:

00
01
02
..

最终你会溢出。

98
99
00

我们有两位数字,可以表示从0到100的所有数字。所有这些数字都是正数!假设我们也想表示负数?

我们真正拥有的是一个循环。2之前的数字是1。1之前的数字是0。0之前的数字是…99.

为了简单起见,我们设任何大于50的数都是负数。0 ~ 49代表0 ~ 49。“99”是-1,“98”是-2,…“50”是-50。

这个表示是十的补数。计算机通常使用2的补码,除了使用位而不是数字之外,它是一样的。

10的补数的好处在于加法运算可以正常进行。你不需要做任何特殊的加法和负数!

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