返回二进制中1的个数
size_t countOf1(int num)
{
size_t count=0;
while(num)
{
num&=num-1;
++count;
}
return count;
}
返回比x大的2的幂级数：
int pow2gt (int x) { --x; x|=x>>1; x|=x>>2; x|=x>>4; x|=x>>8; x|=x>>16; return x+1; }
时间复杂度为O(1)的浮点数开平方根算法：
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;
}
