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

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。

这是一种对负整数进行编码的聪明方法，该方法将数据类型中大约一半的位组合保留给负整数，并且将大多数负整数与其对应的正整数相加会导致进位溢出，使结果为二进制零。

因此，在2的补码中，如果1是0x0001，那么-1是0x1111，因为这将导致0x0000的组合和(溢出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

通过对给定数的第1个补数加1，可以求出两个补数。 假设我们要求出10101的两个补，然后求出它的一个补，也就是，在这个结果上加1，也就是，01010+1=01011，这就是最终答案。

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;