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;
}

Exception up = new Exception("Something is really wrong.");
throw up;  //ha ha

// this is really complicated

没有其他评论了

catch (Ex as Exception)
{
     // oh crap, we should do something.
}

没有什么比一个空的catch块更能让人觉得代码是健壮的....

try {
  dataSource.close();
}
catch (SQLException ex) {
  // Do nothing, since we're going to trash this anyway
}

当然，这类事情实际上是JDBC(或者至少是Oracle的JDBC驱动程序)中的wtf，因为它可以在关闭连接时抛出SQLExceptions…

// The following strings are meant to be funny.  Do not edit these strings
// unless you are funny, too.  If you don't know if you're funny, you're
// not funny.  If fewer than 2 people unrelated to you have told you that 
// you're funny, you're not funny.