位运算
题目:
int func(unsigned int i)
{
Unsigned int temp=i
Temp=(temp & 0x55555555)+((temp & 0xaaaaaaaa)>>1);
Temp=(temp & 0x33333333)+((temp & 0xccccccccc)>>2);
Temp=(temp & 0x0f0f0f0f)+((temp & 0xf0f0f0f0>>4);
Temp=(temp & 0xff00ff)+((temp & 0xff00fff00)>>8);
Temp=(temp & 0xffff)+((temp & 0xffff0000)>>16);
Return temp;
}
网上找到大牛说明:
unsigned int bitcount(register unsigned int x)
{
x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
x = (x & 0x0000FFFF) + ((x >>16) & 0x0000FFFF);
return(x & 0x0000003f);
}
//进一步解释:看图吧,已经山穷水尽啦,如果还不明白请打120
/*版一*/
unsigned int bitcount(unsigned int x)
{
int b;
for(b=0;x != 0;x >>= 1)
{
if(x&01)
b++;
}
return b;
}
/*版二*/ unsigned int bitcount(unsigned int x) { int b; for(b=0;x != 0;) { x &= (x-1); /* 这里的做用是每次都会在原的的数的基础上去掉一个1,贼酷 */ b++; } return b; } 转载自: http://www.cnitblog.com/wuzi/archive/2008/10/05/49851.html