我的多线程发牌程序错在哪里

Question

import java.util.*;enum Color{ 红桃,黑桃,方片,梅花 }
public class TestThread
{
public static void main(String[] args)
{
CardHeap cardheap = new CardHeap();
Player p1 = new Player(cardheap, 1,"A");
Player p2 = new Player(cardheap, 2,"B");
Player p3 = new Player(cardheap, 3,"C");
Player p4 = new Player(cardheap, 4,"D"); p1.start();
p2.start();
p3.start();
p4.start();

}
}class CardHeap
{
LinkedList<Card> cardlist = new LinkedList<Card>();
int turn = 0;
public CardHeap()
{
String[] arr = {"1","2","3","4","5","6","7","8","9","10","J","Q","K"};
for(Color c : Color.values()) //生成牌堆
{
for(int j = 0;j < 13; ++j)
{
//把牌添加到该列表的尾
cardlist.addLast(new Card(arr[j],c));
}
}
Collections.shuffle(cardlist);//乱序
}

public int getlistSize()
{
return cardlist.size();
}
public int getTurn()
{
return turn;
}
public void setTurn(int turn)
{
this.turn = turn;
}

public Card getAndRemoveCard()
{
return (Card)cardlist.poll();
}
}class Player extends Thread
{
int x = 0;
int id;
CardHeap heap;
String playerName; public Player(CardHeap heap,int id,String name)
{
this.heap = heap;
this.id = id;
this.playerName = name;
} public void run()
{
while(true)
{
getcard();
x++;
if(heap.getlistSize() == 0)
{
System.out.println(playerName +"线程进行了"+ x+ "次调用");
return;
}
}
}

public synchronized void getcard()
{
try
{
if(heap.getTurn() + 1 == id)
{
Card c = heap.getAndRemoveCard();
if(c != null)
{
System.out.println(playerName + "取到" + c.getColor() + " " + c.getNum());
heap.setTurn((heap.getTurn() + 1) % 4);
}
}
else
{
wait();
}
notifyAll();
}
catch(Exception e)
{
System.out.println(e);
}
}
}class Card
{
String num;
Color color; public Card(String num, Color color)
{
this.num = num;
this.color = color;
} public String getNum()
{
return num;
} public Color getColor()
{
return color;
}
}运行结果如下：
A取到方片1这样就卡住了。为什么

Answer 1

这个肯定有问题,假如第一个线程开始执行run方法,此时turn=0,发了一张牌,此时的turn为2,调用notifyAll方法之后,当前线程唤醒,而并不是所有的线程都唤醒了,所以下一次执行之后,其他线程处于wiat状态,而此时只有线程1处于非wait状态,但是条件又不成立,所以这个线程也会挂死.

Answer 2

  public synchronized void getcard()
    {
        try
        {
            if(heap.getTurn() + 1 == id)
            {
                Card c = heap.getAndRemoveCard();
                if(c != null)
                {
                    System.out.println(playerName + "取到" + c.getColor() + " " + c.getNum());
                    heap.setTurn((heap.getTurn() + 1) % 4);
                }
            }
            else
            {
                wait();
            }
            notifyAll();
        }
        catch(Exception e)
        {
            System.out.println(e);
        }
    }你这个同步有什么用，又不是共享资源。
把这个放到CardHeap里面，多个线程去访问这个方法。

Answer 3

enum Color{ 红桃,黑桃,方片,梅花 }
public class TestThread
{
    public static void main(String[] args)
    {
        CardHeap cardheap = new CardHeap();
        Player p1 = new Player(cardheap, 1,"A");
        Player p2 = new Player(cardheap, 2,"B");
        Player p3 = new Player(cardheap, 3,"C");
        Player p4 = new Player(cardheap, 4,"D");        p1.start();
        p2.start();
        p3.start();
        p4.start();
        
    }
}
class CardHeap
{
    LinkedList<Card> cardlist = new LinkedList<Card>();
    int turn = 0;
    public CardHeap()
    {
        String[] arr = {"1","2","3","4","5","6","7","8","9","10","J","Q","K"};
        for(Color c : Color.values()) //生成牌堆
        {
            for(int j = 0;j < 13; ++j)
            {
                //把牌添加到该列表的尾
                cardlist.addLast(new Card(arr[j],c));
            }
        }
        Collections.shuffle(cardlist);//乱序
    }
    
    public int getlistSize()
    {
        return cardlist.size();
    }
    public int getTurn()
    {
        return turn;
    }
    public void setTurn(int turn)
    {
        this.turn = turn;
    }
    
    public Card getAndRemoveCard()
    {
        return (Card)cardlist.poll();
    }
}class Player extends Thread
{
    int x = 0;
    int id;
    CardHeap heap;
    String playerName;    public Player(CardHeap heap,int id,String name)
    {
        this.heap = heap;
        this.id = id;
        this.playerName = name;
    }    public void run()
    {
        while(true)
        {
         synchronized(heap){//锁住对象
         getcard();
                x++;
                if(heap.getlistSize() == 0)
                {
                    System.out.println(playerName +"线程进行了"+ x+ "次调用");
                    return;
                }
         }
            
        }
    }
    
    public  void getcard()
    {
        try
        {
            if(heap.getTurn() + 1 == id)
            {
                Card c = heap.getAndRemoveCard();
                if(c != null)
                {
                    System.out.println(playerName + "取到" + c.getColor() + " " + c.getNum());
                    heap.setTurn((heap.getTurn() + 1) % 4);
                }
            }
            else
            {
                yield(); // 切换线程
            }
        }
        catch(Exception e)
        {
            System.out.println(e);
        }
    }
}class Card
{
    String num;
    Color color;    public Card(String num, Color color)
    {
        this.num = num;
        this.color = color;
    }    public String getNum()
    {
        return num;
    }    public Color getColor()
    {
        return color;
    }
}

Answer 4

wait()，notifyAll()用于对象上，而不用于Thread本身。通常wait()用在同步块中，表示释放锁定的共享资源并暂停，等待其他线程使用或处理此资源后notifyAll()，此线程继续执行。LZ的程序中wait()改为yield()或sleep(10)，去掉notifyAll()就正常运行了。