实践JAVA wait(), notify(),sleep方法--一道多线程的面试题

建立三个线程,A线程打印10次A,B线程打印10次B,C线程打印10次C,要求线程同时运行,交替打印10次ABC。

这个问题用Object的wait(),notify()就可以很方便的解决。

public class MyThreadPrinter2 implements Runnable {     
    
    private String name;     
    private Object prev;     
    private Object self;     
    
    private MyThreadPrinter2(String name, Object prev, Object self) {     
        this.name = name;     
        this.prev = prev;     
        this.self = self;     
    }     
    
    @Override    
    public void run() {     
        int count = 10;     
        while (count > 0) {     
            synchronized (prev) {     
                synchronized (self) {     
                    System.out.print(name);     
                    count--;    
                    self.notify();     
                }     
                try {     
                    prev.wait();     
                } catch (InterruptedException e) {     
                    e.printStackTrace();     
                }     
            }     
    
        } 

 

//如果确实下面这段代码,那么会有两个线程无法退出,因为输出A的线程打印10次后,没办法调用notify()通知打印B的线程,以此内存.

synchronized (self){ 

self.notify();

}

    }     
    
    public static void main(String[] args) throws Exception {     
        Object a = new Object();     
        Object b = new Object();     
        Object c = new Object();     
        MyThreadPrinter2 pa = new MyThreadPrinter2("A", c, a);     
        MyThreadPrinter2 pb = new MyThreadPrinter2("B", a, b);     
        MyThreadPrinter2 pc = new MyThreadPrinter2("C", b, c);     
             
             
        new Thread(pa).start();  
        new Thread(pb).start();  
        new Thread(pc).start();    }     
}    

先来解释一下其整体思路,从大的方向上来讲,该问题为三线程间的同步唤醒操作,主要的目的就是ThreadA->ThreadB->ThreadC->ThreadA循环执行三个线程。

为了控制线程执行的顺序,那么就必须要确定唤醒、等待的顺序,所以每一个线程必须同时持有两个对象锁,才能继续执行。

一个对象锁是prev,就是前一个线程所持有的对象锁。还有一个就是自身对象锁。主要的思想就是,为了控制执行的顺序,必须要先持有prev锁,也就前一个线程要释放自身对象锁,再去申请自身对象锁,两者兼备时打印,之后首先调用self.notify()释放自身对象锁,唤醒下一个等待线程,再调用prev.wait()释放prev对象锁,终止当前线程,等待循环结束后再次被唤醒。

运行上述代码,可以发现三个线程循环打印ABC,共10次。程序运行的主要过程就是A线程最先运行,持有C,A对象锁,后释放A,C锁,唤醒B。线程B等待A锁,再申请B锁,后打印B,再释放B,A锁,唤醒C,线程C等待B锁,再申请C锁,后打印C,再释放C,B锁,唤醒A。

看起来似乎没什么问题,但如果你仔细想一下,就会发现有问题,就是初始条件,三个线程按照A,B,C的顺序来启动,按照前面的思考,A唤醒B,B唤醒C,C再唤醒A。

但是这种假设依赖于JVM中线程调度、执行的顺序。具体来说就是,在main主线程启动ThreadA后,需要在ThreadA执行完,在prev.wait()等待时,再切回线程启动ThreadB,ThreadB执行完,在prev.wait()等待时,再切回主线程,启动ThreadC,只有JVM按照这个线程运行顺序执行,才能保证输出的结果是正确的。而这依赖于JVM的具体实现。

考虑一种情况,如下:如果主线程在启动A后,执行A,过程中又切回主线程,启动了ThreadB,ThreadC,之后,由于A线程尚未释放self.notify,也就是B需要在synchronized(prev)处等待,而这时C却调用synchronized(prev)获取了对b的对象锁。这样,在A调用完后,同时ThreadB获取了prev也就是a的对象锁,ThreadC的执行条件就已经满足了,会打印C,之后释放c,及b的对象锁,这时ThreadB具备了运行条件,会打印B,也就是循环变成了ACBACB了。这种情况,可以通过在run中主动释放CPU,来进行模拟。代码如下:

 

public void run() {     
    int count = 10;     
    while (count > 0) {     
        synchronized (prev) {     
            synchronized (self) {     
                System.out.print(name);     
                count--;    
                try{  
                Thread.sleep(1);  
                }  
                catch (InterruptedException e){  
                 e.printStackTrace();  
                }  
                  
                self.notify();     
            }     
            try {     
                prev.wait();     
            } catch (InterruptedException e) {     
                e.printStackTrace();     
            }     
        }     
  
    }

//如果确实下面这段代码,那么会有两个线程无法退出,因为输出A的线程打印10次后,没办法调用notify()通知打印B的线程,以此内存.

synchronized (self){ 

self.notify();

}

     
}     

运行后的打印结果就变成了ACBACB了。为了避免这种与JVM调度有关的不确定性。需要让A,B,C三个线程以确定的顺序启动,最终代码如下:

public class MyThreadPrinter2 implements Runnable {     
    
    private String name;     
    private Object prev;     
    private Object self;     
    
    private MyThreadPrinter2(String name, Object prev, Object self) {     
        this.name = name;     
        this.prev = prev;     
        this.self = self;     
    }     
    
    @Override    
    public void run() {     
        int count = 10;     
        while (count > 0) {     
            synchronized (prev) {     
                synchronized (self) {     
                    System.out.print(name);     
                    count--;    
                    try{  
                    Thread.sleep(1);  
                    }  
                    catch (InterruptedException e){  
                     e.printStackTrace();  
                    }  
                      
                    self.notify();     
                }     
                try {     
                    prev.wait();     
                } catch (InterruptedException e) {     
                    e.printStackTrace();     
                }     
            }     
    
        } 

//如果确实下面这段代码,那么会有两个线程无法退出,因为输出A的线程打印10次后,没办法调用notify()通知打印B的线程,以此内存.

synchronized (self){ 

  self.notify();

}

    }     
    
    public static void main(String[] args) throws Exception {     
        Object a = new Object();     
        Object b = new Object();     
        Object c = new Object();     
        MyThreadPrinter2 pa = new MyThreadPrinter2("A", c, a);     
        MyThreadPrinter2 pb = new MyThreadPrinter2("B", a, b);     
        MyThreadPrinter2 pc = new MyThreadPrinter2("C", b, c);     
             
             
        new Thread(pa).start();  
        Thread.sleep(10);  
        new Thread(pb).start();  
        Thread.sleep(10);  
        new Thread(pc).start();  
        Thread.sleep(10);  
    }     
}    

下面是用concurrent包实现的同样的逻辑代码。

public class MyThreadPrinter2 implements Runnable {     
    
    private String name;     
    private ReentrantLock prev;     
    private ReentrantLock self;     
    private Condition  prevCondition;
    private Condition  selfCondition;
    
    private MyThreadPrinter2(String name, 
            ReentrantLock prev,Condition  prevCondition, 
            ReentrantLock self,Condition  selfCondition) {     
        this.name = name;     
        this.prev = prev;     
        this.self = self;     
        this.prevCondition = prevCondition;
        this.selfCondition = selfCondition;
    }     
      
    public void run() {     
        int count = 10;     
        while (count > 0) {     
            prev.lock();
            self.lock();
            System.out.print(name);     
            count--;    
            selfCondition.signal();
            self.unlock();
            try {
                prevCondition.await();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            finally {
                prev.unlock(); 
            }
        }
        self.lock();
        selfCondition.signal();
        self.unlock();
    }     
    
    public static void main(String[] args) throws Exception {     
        ReentrantLock a = new ReentrantLock();  
        Condition aCondition = a.newCondition();
        ReentrantLock b = new ReentrantLock();     
        Condition bCondition = b.newCondition();
        ReentrantLock c = new ReentrantLock();    
        Condition cCondition = c.newCondition();
        MyThreadPrinter2 pa = new MyThreadPrinter2("A",c,cCondition,a,aCondition);     
        MyThreadPrinter2 pb = new MyThreadPrinter2("B",a,aCondition,b,bCondition);     
        MyThreadPrinter2 pc = new MyThreadPrinter2("C",b,bCondition,c,cCondition);     
             
        new Thread(pa).start();  
        Thread.sleep(10);  
        new Thread(pb).start();  
        Thread.sleep(10);  
        new Thread(pc).start();  
        Thread.sleep(10);  
       }     
}    

 

posted @ 2014-10-06 17:26  princessd8251  阅读(963)  评论(0编辑  收藏  举报