例：0x0001 -> 0x1000
思路一
result = 0x00000000;
resuilt(从左到右)第i位与x(从右到左)第i位进行“或”运算
与0“或”运算的结果为其本身，而result全是0
unsigned int fun_a(unsigned int x)
{
unsigned int result = 0;
for(int i = 0; i < 32; i++) {
result = (result << 1) | (x & 1);
x = x >> 1;
}
return result;
}思路二
先奇偶位交换，然后以2位为一组，组间再奇偶交换，然后4位一组，8位，16位…
unsigned int fun_b(unsigned int x)
{
x = ((x >> 1) & 0x55555555) | ((x << 1) & 0xaaaaaaaa);
x = ((x >> 2) & 0x33333333) | ((x << 2) & 0xcccccccc);
x = ((x >> 4) & 0x0f0f0f0f) | ((x << 4) & 0xf0f0f0f0);
x = ((x >> 8) & 0x00ff00ff) | ((x << 8) & 0xff00ff00);
x = ((x >> 16)& 0x0000ffff) | ((x << 16)& 0xffff0000);
return x;
}
当然，也可以先把高16位和低16位交换，分别再对16位进行8位交换，递归至2位交换