快速计算一个数的平方根及其倒数

源自 http://www.matrix67.com/blog/archives/362

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

  #ifndef Q3_VM
  #ifdef __linux__
    assert( !isnan(y) ); // bk010122 - FPE?
  #endif
  #endif
  return y;
}

/*
================
SquareRootFloat
================
*/
float SquareRootFloat(float number) {
    long i;
    float x, y;
    const float f = 1.5F;

    x = number * 0.5F;
    y  = number;
    i  = * ( long * ) &y;
    i  = 0x5f3759df - ( i >> 1 );
    y  = * ( float * ) &i;
    y  = y * ( f - ( x * y * y ) );
    y  = y * ( f - ( x * y * y ) );
    return number * y;
}

应用了牛顿迭代法求根，有一个神秘的0x5f3759df，因为0x5f3759df – (i >> 1)出人意料地接近根号y的倒数!!!