返回二進位制中1的個數

size_t countOf1(int num)
{
size_t count=0;
while(num)
{
num&=num-1;
++count;
}
return count;
}

返回比x大的2的冪級數：

int pow2gt (int x)	{	--x;	x|=x>>1;	x|=x>>2;	x|=x>>4;	x|=x>>8;	x|=x>>16;	return x+1;	}

時間複雜度為O(1)的浮點數開平方根演算法：

float SqrtByCarmack( float number )
{
    int i;
    float x2, y;
    const float threehalfs = 1.5F;
    x2 = number * 0.5F;
    y  = number;
    i  = * ( int * ) &y;   
    i  = 0x5f375a86 - ( i >> 1 );
    y  = * ( float * ) &i;
    y  = y * ( threehalfs - ( x2 * y * y ) );
    y  = y * ( threehalfs - ( x2 * y * y ) ); 
    y  = y * ( threehalfs - ( x2 * y * y ) );
    return number*y;
}