为什么此代码无法按预期工作

Question

为什么x_temp没有更新值，因为注释行x &= ~(1 << i);工作正常。 哪里出错？

int x = 0x4567;
int x_temp = 0x0;// = 0xF0FF;
int y = 0x0200;
int i;
for(i = 8; i < 12; i++)
{//clean clear
    x_temp = x & ~(1 << i);
    //x &= ~(1 << i); //This line works perfectly.
}
printf("x_temp = %x....\n", x_temp);//Still it retains the value 0x4567.
printf("x = %x....\n", x);
y = x|y; //y = x_temp|y;
printf("y = %x\n", y);

Answer 1

在循环的最后一次迭代中，i为11，但x的第11位已经为0，因此结果为0x4567。我不知道你为什么期待别的东西。在x &= ~(1 << i)的情况下，您在x的之前值中清除一点，而使用x_temp时，您会为x_temp分配一个新值} ...一个案例是累积的，另一个案例不是。

考虑两个循环的痕迹：

for `x &= ~(1 << i)`, you have
x is 0x4567 originally
x is 0x4467 after clearing 1<<8
x is 0x4467 after clearing 1<<9
x is 0x4067 after clearing 1<<10
x is 0x4067 after clearing 1<<11

     

但

for `x_temp = x & ~(1 << i)`, you have
x is 0x4567 (originally and forever)
x_temp is 0x4467 after clearing 1<<8 from x (which hasn't changed)
x_temp is 0x4567 after clearing 1<<9 from x (which hasn't changed)
x_temp is 0x4167 after clearing 1<<10 from x (which hasn't changed)
x_temp is 0x4567, after clearing 1<<11 from x (which hasn't changed)

也许这更清楚： 假设x = 5;那么设置x + = 1的循环将产生6,7,8,9,10的值，...... 但设置x_temp = x + 1的循环将产生值6,6,6,6,6，...

Answer 2

也许是因为您丢弃了x_temp的旧值？

for(i = 8; i < 12; i++)
{
    x_temp = x & ~(1 << i);
}

与

相同

x_temp = x & ~(1 << 11);