static int SQRT (int nRoot)
{
int nSqrt = 0;
for(int i = 0x10000000; i != 0; i >>= 2)
{
int nTemp = nSqrt + i;
nSqrt >>= 1;
if(nTemp <= nRoot)
{
nRoot -= nTemp;
nSqrt += i;
}
}
return nSqrt;
}
用到了 (a+b)(a+b) = a*a + b*b + 2ab
测试了下, 由于 for 循环执行完成需16次循环。因此, 无论给定什么数, 都在16步给出结果。
Another way:
/*
** float q_rsqrt( float number )
*/
float q_rsqrt( float number )
{
long i;
float x2, y;
const float threehalfs = 1.5F;
x2 = number * 0.5F;
y = number;
i = * ( long * ) &y; // evil floating point bit level hacking
i = 0x5f3759df - ( i >> 1 ); // what the fuck?
y = * ( float * ) &i;
y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration
// y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed
#ifdef __linux__
assert( !isnan(y) ); // bk010122 - FPE?
#endif
return y;
}