Question

这是一种C拼图。您必须判断程序是否完成执行，如果是，运行所需的时间以及返回操作系统的内容。

static unsigned char buffer[256];

int main(void)
{
  unsigned char *p, *q;
  q = (p = buffer) + sizeof(buffer);
  while (q - p)
  {     
      p = buffer;
      while (!++*p++);
  }
  return p - q;
}

[编辑] 我删除了面试问题标签，因为这似乎是人们反对的主要内容。这是一个很棒的小谜题，但正如大家已经指出的那样，这不是一个很好的面试问题。

Answer 1

尽管这是一个可怕的采访问题，但实际上非常有趣：

static unsigned char buffer[256];

int main(void)
{
  unsigned char *p, *q;
  q = (p = buffer) + sizeof(buffer);
  /* This statement will set p to point to the beginning of buffer and will
     set q to point to one past the last element of buffer (this is legal) */
  while (q - p)
  /* q - p will start out being 256 and will decrease at an inversely 
     exponential rate: */
  {     
      p = buffer;
      while (!++*p++);
      /* This is where the interesting part comes in; the prefix increment,
         dereference, and logical negation operators all have the same
         precedence and are evaluated **right-to-left**.  The postfix
         operator has a higher precedence.  *p starts out at zero, is
         incremented to 1 by the prefix, and is negated by !.
         p is incremented by the postfix operator, the condition
         evaluates to false and the loop terminates with buffer[0] = 1.

         p is then set to point to buffer[0] again and the loop continues 
         until buffer[0] = 255.  This time, the loop succeeds when *p is
         incremented, becomes 0 and is negated.  This causes the loop to
         run again immediately after p is incremented to point to buffer[1],
         which is increased to 1.  The value 1 is of course negated,
         p is incremented which doesn't matter because the loop terminates
         and p is reset to point to buffer[0] again.

         This process will continue to increment buffer[0] every time,
         increasing buffer[1] every 256 runs.  After 256*255 runs,
         buffer[0] and buffer[1] will both be 255, the loop will succeed
         *twice* and buffer[2] will be incremented once, etc.

         The loop will terminate after about 256^256 runs when all the values
         in the buffer array are 255 allowing p to be incremented to the end
         of the array.  This will happen sometime after the universe ends,
         maybe a little sooner on the new Intels ;)
      */
  }
  return p - q;
  /* Returns 0 as p == q now */
}

本质上这是256位（假设8位字节）计数器，256位数，当整个计数器“翻转”时，程序将退出。

这很有趣的原因是因为代码实际上是完全合法的C（在这些类型的问题中通常没有找到未定义的或实现定义的行为）并且实际上存在合法的算法问题，尽管有点隐藏在混合。这是一个可怕的面试问题的原因是因为我不希望任何人记住while语句中涉及的运算符的优先级和关联性。但它确实可以带来有趣且富有洞察力的小练习。

Answer 2

此代码是垃圾，请参阅注释

static unsigned char buffer[256];
int main(void)
{
  unsigned char *p, *q;
  q = (p = buffer) + sizeof(buffer);    //p=buffer, q=buffer+256
  while (q - p)    //q-p = 256 on first iteration
  {     
      p = buffer;        //p=buffer again
      while (!++*p++);   //increment the value pointed at by p+1 and check for !0
  }
  return p - q;    //will return zero if loop ever terminates
}

可能会终止，也可能不会; while循环实质上是扫描未初始化的缓冲区，因此它可能会抛出访问冲突;我不记得++ * p ++的绑定优先级，我也不在乎查找它

如果这真的是一个面试问题，我的回答是“如果这是你希望我合作的那种代码，我不想要这份工作”

编辑：感谢Robert Gamble提醒我静态数组自动初始化为零，因此代码不是完全垃圾 - 但我仍然不想维护它或使用编写它的nutjob; - ）< / p>

Answer 3

这个问题的正确答案是：

此代码不可维护，不可测试，无用，应删除或重写。

其他任何事情都意味着受访者不会将其视为软件工程师。

再一次，你可能不会接受工程职位的面试。

Answer 4

unsigned char *p, *q;

这个问题在很多层面都不是吗？首先，是否存在无符号字符？第二，我可能在这里错了，所以不要引用我，但不是char * p， q产生时髦的结果？就是这样，或者它很容易做char p，q，这将是不好的形式。

以下情况要好得多：

char* p;
char* q;

这个程序的输出是什么，它返回操作系统的是什么？

4 个答案: