Question

我已经完成了dining philosophers解决方案。由于在某些时候，典型的监视器实现会导致饥饿。在here

中，我已经完成了“礼貌”版本的Dining-philosopher's问题

monitor diningPhilosophers {
   int[] state = new int[5];
   boolean[] leftHungry = new boolean[5];
   boolean[] rightHungry = new boolean[5];
   static final int THINKING = 0;
   static final int HUNGRY = 1;
   static final int EATING = 2;
   condition[] self = new condition[5];

   public diningPhilosophers {
      for (int i=0;i<5;i++) {
         state[i] = THINKING;
         leftHungry[i] = false;
         rightHungry[i] = false;
      }  
   }   

   public entry pickUp(int i) {
      state[i] = HUNGRY;
      test(i);
      if (state[i] != EATING)
         self[i].wait;
      rightHungry(left(i)) = false;
      leftHungry(right(i)) = false;
   }

   ublic entry putDown(int i) {
      state[i] = THINKING;
      test(left(i));
      if (state[left(i)] == HUNGRY)
         leftHungry[i] = true;
      test(right(i));
      if (state[right(i)] == HUNGRY)
         rightHungry[i] = true;
   }

   private test(int i) {
      if  (state[right(i] != EATING) &&
          (state[i] == HUNGRY) &&
          (state[left(i)] != EATING) &&
             !leftHungry(i) && !rightHungry(i) ) {
             state[i] = EATING;
             self[i].signal;
          }
   }

   private int left(int i) {
      return (i+1)%5;
   }    

   private int right(int i) {
      return (i+4)%5;
   }    

}

那么如果两个相邻的哲学家同时饿着怎么办呢？由于测试（i）正在检查其左右哲学家是否感到饥饿。如果它发现它的相邻的一个也饿了。这有点像僵局吗？我的意思是他们两个都不能吃饭，因为他们相邻的哲学家都饿了吗？

Answer 1

你是对的。如果两个相邻的哲学家在彼此的眼中感到饥饿，就会发生僵局。 leftHungry[0] == TRUE && rightHungry[1] == TRUE。

但，此状态永远不会发生，因为leftHungry[x]或rightHungry[x]设置为TRUE的唯一时间是x调用{{ 1}}，始终以putDown()调用x开头，将pickUp()和leftHungry[x]设置为rightHungry[x]。

例如： FALSE有philosopher0，状态为leftHungry[0] == TRUE，HUNGRY调用philosopher1，其状态为那么pickUp()也是。即使HUNGRY必须等待，（因为philosopher1是philosopher2），EATING设置philosopher0，也阻止任何相邻的哲学家同时思考彼此都饿了。

用餐哲学家的解决方案中的饥饿

1 个答案: