根据这个问答,我试着验证了回答,所以我写了以下内容:
#include <stdio.h>
int main ()
{
int t;int i;
for (i=120;i<140;i++){
t = (i - 128) >> 31;
printf ("t = %X , i-128 = %X , ~t & i = %X , ~t = %X \n", t, i-128 , (~t &i), ~t);
}
return 0;
}
输出结果如下:
t = FFFFFFFF , i-128 = FFFFFFF8 , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFF9 , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFA , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFB , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFC , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFD , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFE , ~t & i = 0 , ~t = 0
t = FFFFFFFF , i-128 = FFFFFFFF , ~t & i = 0 , ~t = 0
t = 0 , i-128 = 0 , ~t & i = 80 , ~t = FFFFFFFF
t = 0 , i-128 = 1 , ~t & i = 81 , ~t = FFFFFFFF
t = 0 , i-128 = 2 , ~t & i = 82 , ~t = FFFFFFFF
t = 0 , i-128 = 3 , ~t & i = 83 , ~t = FFFFFFFF
t = 0 , i-128 = 4 , ~t & i = 84 , ~t = FFFFFFFF
t = 0 , i-128 = 5 , ~t & i = 85 , ~t = FFFFFFFF
t = 0 , i-128 = 6 , ~t & i = 86 , ~t = FFFFFFFF
t = 0 , i-128 = 7 , ~t & i = 87 , ~t = FFFFFFFF
t = 0 , i-128 = 8 , ~t & i = 88 , ~t = FFFFFFFF
t = 0 , i-128 = 9 , ~t & i = 89 , ~t = FFFFFFFF
t = 0 , i-128 = A , ~t & i = 8A , ~t = FFFFFFFF
t = 0 , i-128 = B , ~t & i = 8B , ~t = FFFFFFFF
为什么在整数变量的情况下,任何负数的~t都等于-1 == 0xFFFFFFFF?
~0 == 0xFFFFFFFF
的意思是对数字0进行按位取反操作,得到的结果为十六进制数0xFFFFFFFF。~~x==x
的意思是对于任何整数x,对其进行两次按位取反操作,结果仍为原来的x。你只需翻转0和-1这两个数字的二进制位。假设使用32位整数。 - Aneri