java Implement Runnable未显示正确的结果

Question

我试图对一个类使用wait（）和notify（）来通知等待的线程，但情况是，谁先进入同步块，谁就会等待其他线程打印一些值并通知等待的线程。但等待的线索是继续等待。我做错了什么

public class MyRunnableClass implements Runnable {
public static boolean transfer = false;
public static boolean isAnyThreadWaiting = false;

int a=10;
@Override
public void run() {
    // TODO Auto-generated method stub

    synchronized(this) {    
        System.out.println("I am Thread : "+Thread.currentThread().getName()+" and I first occuipied block");
        
        try {
            if(isAnyThreadWaiting)
                MyRunnableClass.transfer=true;

            if(!MyRunnableClass.transfer&&!isAnyThreadWaiting) {
                System.out.println("I am Thread : "+Thread.currentThread().getName()+" and I am going to wait");
                while(!MyRunnableClass.transfer) {
                    isAnyThreadWaiting=true;
                    this.wait();
                }
                
            }
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        
        for(int i=0; i<a;i++) {
            System.out.println("I am Thread : "+Thread.currentThread().getName()+" Value of  a is : "+i );
            
        }
        a=0;
        isAnyThreadWaiting=false;
        
        this.notify();
    
        
    }
    
}

public static void main(String [] args) throws InterruptedException {
MyRunnableClass runnable1= new MyRunnableClass();
Thread thread1=new Thread(runnable1,"t1");
Thread thread2=new Thread(runnable1,"t2");
    thread1. start();
    
    
    
    thread2. start();
    }
}

Answer 1

synchronized(this)

this在两个线程中都是相同的，因为您使用了相同的Runnable实例^{}确保两个线程不能同时在块内。这意味着一个线程位于synchronized块的开头，另一个线程位于wait。您需要将同步范围缩小到实际使用共享状态的位置，即仅synchronized(this)实际引用（读取或写入）两个线程所需变量的代码

Answer 2

对一个线程中的静态变量所做的更改对另一个线程不可见。因此，对于这两个线程，它总是transfer==false和isAnyThreadWaiting==false

这意味着两个线程都会进入“wait（）”块并永远卡在那里

因为您试图对静态变量进行同步，所以必须使用synchronized static方法

请参见此处的更多详细信息： How to synchronize a static variable among threads running different instances of a class in Java?

但更简单的解决方案是根本不使用静态变量。 在您的情况下，两个线程使用同一个对象，因此不需要将变量设置为静态：

private boolean transfer = false;
private boolean isAnyThreadWaiting = false;

完整的解决方案应该如下所示：

public class MyRunnableClass implements Runnable {
private boolean transfer = false;
private boolean isAnyThreadWaiting = false;

int a=10;

@Override
public void run() {
    // TODO Auto-generated method stub

    synchronized(this) {    
        System.out.println("I am Thread : "+Thread.currentThread().getName()+" and I first occuipied block");
        
        try {
            if(isAnyThreadWaiting)
                transfer=true;

            if(!transfer && !isAnyThreadWaiting) {
                System.out.println("I am Thread : "+Thread.currentThread().getName()+" and I am going to wait");
                while(!transfer) {
                    isAnyThreadWaiting=true;
                    this.wait();
                }
                
            }
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        
        for(int i=0; i<a;i++) {
            System.out.println("I am Thread : "+Thread.currentThread().getName()+" Value of  a is : "+i );
            
        }
        a=0;
        isAnyThreadWaiting=false;
        
        this.notify();
    
        
    }
    
}

public static void main(String [] args) throws InterruptedException {
MyRunnableClass runnable1= new MyRunnableClass();
Thread thread1=new Thread(runnable1,"t1");
Thread thread2=new Thread(runnable1,"t2");
    thread1. start();
    
    
    
    thread2. start();
    }
}

Answer 3

我已经调试了你的代码并找到了原因：

a=0;

请删除此代码，您将看到两个线程的输出：

I am Thread : t1 and I first occuipied block
I am Thread : t1 and I am going to wait
I am Thread : t2 and I first occuipied block
I am Thread : t2 Value of  a is : 0
I am Thread : t2 Value of  a is : 1
I am Thread : t2 Value of  a is : 2
I am Thread : t2 Value of  a is : 3
I am Thread : t2 Value of  a is : 4
I am Thread : t2 Value of  a is : 5
I am Thread : t2 Value of  a is : 6
I am Thread : t2 Value of  a is : 7
I am Thread : t2 Value of  a is : 8
I am Thread : t2 Value of  a is : 9
I am Thread : t1 Value of  a is : 0
I am Thread : t1 Value of  a is : 1
I am Thread : t1 Value of  a is : 2
I am Thread : t1 Value of  a is : 3
I am Thread : t1 Value of  a is : 4
I am Thread : t1 Value of  a is : 5
I am Thread : t1 Value of  a is : 6
I am Thread : t1 Value of  a is : 7
I am Thread : t1 Value of  a is : 8
I am Thread : t1 Value of  a is : 9

另外，示例中的第二个线程实际上没有进入wait（）状态

但是，由于通过第二个线程将a设置为0，第一个线程跳过了这个周期：

for(int i=0; i<a;i++) {
    System.out.println("I am Thread : "+Thread.currentThread().getName()+" Value of  a is : "+i );
    
}

这就是为什么你看不到第一个线程的输出。但它并没有停留在等待（）状态