/*
 * You may think you know what the following code does.
 * But you dont. Trust me.
 * Fiddle with it, and youll spend many a sleepless
 * night cursing the moment you thought youd be clever
 * enough to "optimize" the code below.
 * Now close this file and go play with something else.
 */ 

public int hashCode() {
//sucks, but what're you gonna do

/*
int hash = 7;
for (int i = 0; i < array.length; i++)
    hash = hash * 31 * (null == array[i] ? 0 : array[i].hashCode());
return hash;
*/

return 0;
}

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;
}

大概30页的xslt的中间部分

<!-- Here be dragons  -->

// I'm sorry.

//ALL YOUR BASE ARE BELONG TO US

.．.这让我的老板以为有人黑进了系统。他不知道这个笑话。