问题描述
我想知道这句话背后的逻辑，即证明.对于任何x，C表达式-x，〜x + 1和〜(x-1)都产生相同的结果.对于特定示例，我可以证明这是正确的.我认为证明这一点的方式与二进制补码的性质有关.有任何想法吗?
I want to know the logic behind this statement, the proof. The C expression -x, ~x+1, and ~(x-1) all yield the same results for any x. I can show this is true for specific examples. I think the way to prove this has something to do with the properties of two's complement. Any ideas?
推荐答案
请考虑将数字按位补码时所得到的结果.n位整数x的按位补码在所有x都为0的地方都为1，反之亦然.因此很明显可以看到:
Consider what you get when you add a number to its bitwise complement. The bitwise complement of an n-bit integer x has a 1 everywhere x has a 0, and vice versa. So it's clear to see:
x +〜x = 0b11 ... 11(全1的n位值)
x + ~x = 0b11...11 (n-bit value of all ones)
无论x中的位数如何.此外，请注意，在一个全为1的n位数字上加一个将使其换为零.因此，我们看到:
Regardless of the number of bits in x. Further, note that adding one to an n-bit number filled with all ones will make it wrap to zero. Thus we see:
x +〜x + 1 = 0b11 ... 11 + 1 = 0和〜x +1 = -x.
x + ~x + 1 = 0b11...11 + 1 = 0and ~x + 1 = -x.
类似地，注意(x-1)+〜(x-1)= 0b11 ... 11.然后(x-1)+〜(x-1)+1 = 0，〜(x-1)= -x.
Similarly, note (x - 1) + ~(x - 1) = 0b11...11. Then (x - 1) + ~(x - 1) + 1 = 0, and ~(x - 1) = -x.
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