设法把这个糟糕的双关语插入到我们的代码中

for (bo_thans = 0 ; bo_thans < MAX ; bo_thans++)
{
    if(rs == thing[bo_thans])
    {
       found = true;
    }
}

if(!found)
{
   /* Failed to find rs with bo_thans */
   ...
}

早在80年代早期，我在汇编器中看到了这个(引用于模糊的记忆):

I don't understand how the following bit works, but it worked in the program I stole it from.

// Iced odnako
bool Iced{get;set;}

options.BatchSize = 300; //Madness? THIS IS SPARTA!

// If you're reading this, that means you have been put in charge of my previous project.
// I am so, so sorry for you. God speed.

从《雷神之锤III》的资料中，我在一些随机的帖子中偶然发现了这一点。该文件的完整源代码可以在这里找到。这是一种非常快速的求平方根倒数的方法。至于最好的评论呢?当然，这是一种常见的方法，但考虑到它附着在直线上，它的神奇之处在于它的伟大之处。

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;
}