Question

嗨，我正在研究睡觉的理发师问题。当他们到达时增加了优先顾客，他们走在前面，他们是下一个理发师。

我正在使用linkedlist，如果我看到优先客户，我将他列入列表的开头，如果客户不是优先考虑，他会到列表的末尾。然后我调用wantHaircut方法获取列表的第一个元素。

我的问题是客户按照他们到达的顺序进行处理，优先客户必须等待。这是所有发生的代码。我的错误是什么？感谢

 public void arrivedBarbershop(Customer c){

      if(waiting < numChairs && c.isPriority()){
          System.out.println("Customer " + c.getID() + ": is a priority customer - SITTING -");
          mutex.up();             
          customer_list.addFirst(c);
      }
      else if(waiting >= numChairs && c.isPriority()){
          System.out.println("Customer " + c.getID() + ": is a priority customer - STANDING -");
          mutex.up();
          customer_list.addFirst(c);
      } 
      else if(waiting < numChairs && !c.isPriority()){
           waiting++;
           System.out.println("Customer " + c.getID()  + ": arrived, sitting in the waiting room");
          customer_list.addLast(c);
           customers.up(); // increment waiting customers

       }
       else if(waiting >= numChairs && !c.isPriority()) {

         System.out.println("Customer " + c.getID() + ": went to another barber because waiting room was full - " + waiting + " waiting");
         mutex.up();
       }

      if(!customer_list.isEmpty()){
         this.wantHairCut(customer_list.removeFirst());
      }

   }

   public void wantHairCut(Customer c) {
          mutex.up();
          barber.down();  // waits for being allowed in barber chair
          System.out.println("Customer " + c.getID() + ": getting haircut");
          try {
             /** haircut takes between 1 and 2 seconds **/
              Thread.sleep(Barbershop.randomInt(1, 2) * 1000);
          } catch (InterruptedException e) { }
          System.out.println("Barber: finished cutting customer " + c.getID() + "'s hair");
          c.gotHaircut = true;          
          cutting.up();   // signals cutting has finished

          /** customer must pay now **/
          this.wantToCashout(c);          
   }

Answer 1

您必须使用同步集合：

Collections.synchronizedList(List<Object> list)

此方法返回基于参数列表的同步List实例

Answer 2

我很确定逻辑错误就在这里：

  if(waiting < numChairs && c.isPriority()){
      System.out.println("Customer " + c.getID() + ": is a priority customer - SITTING -");
      mutex.up();             
      customer_list.addFirst(c);
  }
  else if(waiting >= numChairs && c.isPriority()){
      System.out.println("Customer " + c.getID() + ": is a priority customer - STANDING -");
      mutex.up();
      customer_list.addFirst(c);
  }

是的，避免复制/粘贴编程！

Answer 3

无论何时修改列表，都应该将该代码放在同步块上。

一种方法：

synchronized(customer_list) {
 customer_list.addFirst(c);
}

另一个，所以你不必在整个地方编码：

public synchronized void addCustomerAtStart(Customer c) {
 customer_list.addFirst(c);
}

然后替换该功能。

当你执行customer.up（）时，看起来它也需要同步，否则你可能会遇到很多不一致的行为。

你甚至不需要使用“mutex.up（）” 提示：如果必须同步线程，您可能需要查看CountDownLatch，它非常易于使用，可以稳定地用于线程同步。

Answer 4

您的代码会将客户添加到列表中（开始或结束），然后从列表中删除客户。该列表最多只包含一个客户，因此客户按其添加顺序进行处理。

Java线程无法与linkedlist一起正常工作

4 个答案: