一个多线程分别打印abc的问题

[welen]

运行这段代码，前面可以打印两三次abc，然后就卡住了，初步估计应该都进入wait了，但是不明白为什么，求指导

public class TestDemo {

	public static void main(String[] args) {
		final Locker p = new Locker();
		Task11 a = new Task11(p);
		Task21 b = new Task21(p);
		Task31 c = new Task31(p);
		Thread one = new Thread(a);
		one.start();
		Thread two = new Thread(b);
		two.start();
		Thread three = new Thread(c);
		three.start();
	}
}


class Task11 implements Runnable {
	
	private Locker obj;
	public Task11(Locker p){
		this.obj = p;
	}
	
	@Override
	public void run() {
		for(int i = 0; i < 5; i++){
			synchronized (obj) {
				if(obj.num != 0){
					try {
						obj.wait();
					} catch (InterruptedException e) {
						e.printStackTrace();
					}
				}else{
					System.out.println("a");
					obj.num = 1;
					obj.notifyAll();
				}
			}
		}
	}
}
class Task21 implements Runnable {
	
	private Locker obj;
	public Task21(Locker p){
		this.obj = p;
	}
	
	@Override
	public void run() {
		for(int i = 0; i < 5; i++){
			synchronized (obj) {
				if(obj.num != 1){
					try {
						obj.wait();
					} catch (InterruptedException e) {
						e.printStackTrace();
					}
				}else{
					System.out.println("b");
					obj.num = 2;
					obj.notifyAll();
				}
			}
		}
	}
}
class Task31 implements Runnable {
	
	private Locker obj;
	public Task31(Locker p){
		this.obj = p;
	}
	
	@Override
	public void run() {
		for(int i = 0; i < 5; i++){
			synchronized (obj) {
				if(obj.num != 2){
					try {
						obj.wait();
					} catch (InterruptedException e) {
						e.printStackTrace();
					}
				}else{
					System.out.println("c");
					obj.num = 0;
					obj.notifyAll();
				}
			}
		}
	}
}

class Locker {
	int num = 0;
}

[faker-_-]

测试了一下，每个线程都去掉else就可以了。因为假如A线程在sleep被唤醒时，它只会进行执行if下面的程序，而if下面没有唤醒其他线程的代码，这样所有的代码都是出于sleep状态了。

[tony4geek]

one two three 的调度顺序问题。

[welen]

引用 1 楼 rui888 的回复:

one two three 的调度顺序问题。

怎么改，按我理解，每次三个线程中总有一个执行了else的代码块，然后通过notifyall唤醒所有等待的，下一次应该就只有num指定的线程可以执行else的代码了。。这样每次都可以有一个else可以得到执行。。事实不是这样，所以不理解了，能说清楚吗

[tony4geek]

你理解下

public class TestDemo {
	 public static void main(String[] args) {
	        final Locker p = new Locker();
	        Task11 a = new Task11(p);
	        Task21 b = new Task21(p);
	        Task31 c = new Task31(p);
	        Thread one = new Thread(a);
	        one.start();
	        Thread two = new Thread(b);
	        two.start();
	     //   Thread three = new Thread(c);
	     //   three.start();
	    }
	}
	 
	 
	class Task11 implements Runnable {
	     
	    private Locker obj;
	    public Task11(Locker p){
	        this.obj = p;
	    }
	     
	    @Override
	    public void run() {
	        for(int i = 0; i < 5; i++){
	            synchronized (obj) {
	                if(obj.num != 0){
	                    try {
	                        obj.wait();
	                    } catch (InterruptedException e) {
	                        e.printStackTrace();
	                    }
	                }else{
	                    System.out.println("a");
	                    obj.num = 1;
	                    obj.notifyAll();
	                }
	            }
	        }
	    }
	}
	class Task21 implements Runnable {
	     
	    private Locker obj;
	    public Task21(Locker p){
	        this.obj = p;
	    }
	     
	    @Override
	    public void run() {
	        for(int i = 0; i < 5; i++){
	            synchronized (obj) {
	                if(obj.num != 1){
	                    try {
	                        obj.wait();
	                    } catch (InterruptedException e) {
	                        e.printStackTrace();
	                    }
	                }else{
	                    System.out.println("b");
	                    obj.num = 0;
	                    obj.notifyAll();
	                }
	            }
	        }
	    }
	}

	class Task31 implements Runnable {
	     
	    private Locker obj;
	    public Task31(Locker p){
	        this.obj = p;
	    }
	     
	    @Override
	    public void run() {
	        for(int i = 0; i < 5; i++){
	            synchronized (obj) {
	                if(obj.num != 2){
	                    try {
	                        obj.wait();
	                    } catch (InterruptedException e) {
	                        e.printStackTrace();
	                    }
	                }else{
	                    System.out.println("c");
	                    obj.num = 0;
	                    obj.notifyAll();
	                }
	            }
	        }
	    }
	}
	 
	class Locker {
	    int num = 0;
	}

[nmyangym]

我的粗浅看法： 楼主这样的逻辑，用while 好一些。看下面的代码：

public class TestDemo {
	 
    public static void main(String[] args) {
        final Locker p = new Locker();
        Task11 a = new Task11(p);
        Task21 b = new Task21(p);
        Task31 c = new Task31(p);
        Thread one = new Thread(a);
        one.start();
        Thread two = new Thread(b);
        two.start();
        Thread three = new Thread(c);
        three.start();
    }
}
 
 
class Task11 implements Runnable {
     
    private Locker obj;
    public Task11(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                while(obj.num != 0){//if 改成while,只要num != 0, 肯定等待。
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                //必须num == 0 ， 才执行。
                System.out.println("a");
                obj.num = 1;
                obj.notifyAll();
            }
        }
    }
}
class Task21 implements Runnable {
     
    private Locker obj;
    public Task21(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                while(obj.num != 1){
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println("b");
                obj.num = 2;
                obj.notifyAll();
            }
        }
    }
}
class Task31 implements Runnable {
     
    private Locker obj;
    public Task31(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                while(obj.num != 2){
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                System.out.println("c");
                obj.num = 0;
                obj.notifyAll();
            }
        }
    }
}
 
class Locker {
    int num = 0;
}

如果不用while, 需要这样改：

public class TestDemo {
	 
    public static void main(String[] args) {
        final Locker p = new Locker();
        Task11 a = new Task11(p);
        Task21 b = new Task21(p);
        Task31 c = new Task31(p);
        Thread one = new Thread(a);
        one.start();
        Thread two = new Thread(b);
        two.start();
        Thread three = new Thread(c);
        three.start();
    }
}
 
 
class Task11 implements Runnable {
     
    private Locker obj;
    public Task11(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                if(obj.num != 0){
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    //唤醒时判断num值，是0，执行输出；否则，本次唤醒无意义，需要把i减一，继续循环。而用while 就比较好。楼主试一下。
                    if(obj.num == 0){
                    	System.out.println("a");
                        obj.num = 1;
                        obj.notifyAll();
                    }
                    else {
                    	i--;
                    }
                }
                else {
                	System.out.println("a");
                    obj.num = 1;
                    obj.notifyAll();
                }
            }
        }
    }
}
class Task21 implements Runnable {
     
    private Locker obj;
    public Task21(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                if(obj.num != 1){
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    if(obj.num == 1){
                    	System.out.println("b");
                        obj.num = 2;
                        obj.notifyAll();
                    }
                    else {
                    	i--;
                    }
                }
                else {
                	System.out.println("b");
                    obj.num = 2;
                    obj.notifyAll();
                }
            }
        }
    }
}
class Task31 implements Runnable {
     
    private Locker obj;
    public Task31(Locker p){
        this.obj = p;
    }
     
    @Override
    public void run() {
        for(int i = 0; i < 5; i++){
            synchronized (obj) {
                if(obj.num != 2){
                    try {
                        obj.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                    if(obj.num == 2){
                    	System.out.println("c");
                        obj.num = 0;
                        obj.notifyAll();
                    }
                    else {
                    	i--;
                    }
                }
                else {
                	System.out.println("c");
                    obj.num = 0;
                    obj.notifyAll();
                }
            }
        }
    }
}
 
class Locker {
    int num = 0;
}

[陆荃]

问题分析： for循环可能会执行到wait分支，故不可能打印所有的5个a（或者b，c）。 至于程序有时（占大比例）会停住，是因为比如task11已经跑完for循环，而此时task21,31均处于wait状态，没有地方触发notifyAll，所以会停住。 解决方法： 问题关键点在于for循环的5次肯定不能输出五次a，b，c。 愚以为，作如下改动能实现楼主要求： public class TestDemo { public static void main(String[] args) { final Locker p = new Locker(); Task11 a = new Task11(p); Task21 b = new Task21(p); Task31 c = new Task31(p); Thread one = new Thread(a, "Task11"); one.start(); Thread two = new Thread(b, "Task21"); two.start(); Thread three = new Thread(c, "Task31"); three.start(); } } class Task11 implements Runnable { private Locker obj; public Task11(Locker p) { this.obj = p; } @Override public void run() { int i = 0; while (true) { synchronized (obj) { if (obj.num != 0) { try { obj.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } else { System.out.println("a"); obj.num = 1; obj.notifyAll(); if (4 == i++) break; } } } } } class Task21 implements Runnable { private Locker obj; public Task21(Locker p) { this.obj = p; } @Override public void run() { int i = 0; while (true) { synchronized (obj) { if (obj.num != 1) { try { obj.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } else { System.out.println("b"); obj.num = 2; obj.notifyAll(); if (4 == i++) break; } } } } } class Task31 implements Runnable { private Locker obj; public Task31(Locker p) { this.obj = p; } @Override public void run() { int i = 0; while (true) { synchronized (obj) { if (obj.num != 2) { try { obj.wait(); } catch (InterruptedException e) { e.printStackTrace(); } } else { System.out.println("c"); obj.num = 0; obj.notifyAll(); if (4 == i++) break; } } } } } class Locker { int num = 0; } 浅薄之见，还望指教。