c - 这个程序的输出是什么，它返回给操作系统的是什么？

Question

这有点像 C 谜题。你必须告诉程序是否完成了它的执行，如果是，它运行了多少时间以及它返回给操作系统的是什么。

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]我删除了采访问题标签，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，但正如每个人都已经指出的那样，这不是一个很好的面试问题。

Answer 1

尽管这是一个可怕的面试问题，但它实际上很有趣:

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 */
}

本质上这是一个 256 位的 base-256(假设 8 位字节)计数器，当整个计数器“翻转”时程序将退出。

之所以有趣，是因为代码实际上是完全合法的 C(没有您通常在这些类型的问题中发现的未定义或实现定义的行为)并且实际上有一个合法的算法问题，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。但它确实是一个有趣且富有洞察力的小练习。