这是我最喜欢的评论。

/// I intend to do this as shittily as possible because there are many better products that will totally blow this out of the water
/// and we don't have them so whatever

在后面的文件中，我们有更多的乐趣

/// sidestep a bug in WCF (that we can't send types across)
/// or, depending on how you look at, this issue is a Feature

稍后再讲

if( where == null)//be nice

不是在代码中，而是在相关的bug跟踪系统中:

这不可能是我代码中的bug。我非常小心地编码。

这是在PHP代码中找到的

$s=2; // chicken and bacon wrap for lunch

多么有用，幸运的是$s是不言自明的

从一个遗留的Perl CGI脚本:

# This is convoluted and evil, sorry.

// long live COM'n'Roll
public enum StatusCode
{
        //success codes
        S_OK                                            = 1,
        S_NONE                                          = 2,
        S_SQL_OPERATIONS_LISTS_EMPTY                    = 3,

        //error codes
        E_NO_MATCHING_END_FOUND                         = -1,
        E_SEQUENCE_NUMBER_NOT_FOUND_AT_BEGINNING        = -2,
        E_SEQUENCE_NUMBER_NOT_FOUND_AT_END              = -3,
        E_FORWARD_AND_BACKWARD_OPS_COUNT_DO_NOT_MATCH   = -4,
        E_FORWARD_AND_BACKWARD_IDS_DO_NOT_MATCH         = -5,
        E_IDS_DO_NOT_MATCH                              = -6
}

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;
}