java - volatile 关键字: Need explanation for this program behaviour

Question

通过该程序理解 volatile 关键字:

public class VolatileExample implements Runnable{    
private volatile int vol;
@Override
public void run(){
    vol=5;
    while(vol==5)
    {
        System.out.println("Inside run when vol is 5. Thread is : "+Thread.currentThread().getName());
            if("t2".equals(Thread.currentThread().getName()))
            {
                System.out.println("Got Thread : "+Thread.currentThread().getName()+" Now Calling Stop To End This Flow");
                stopIt();
            }
    }
}
public void stopIt(){
    vol=10;
}

public static void main(String[] args){
     VolatileExample ve1 = new VolatileExample();
     VolatileExample ve2 = new VolatileExample();

     Thread t1 = new Thread(ve1);
     Thread t2 = new Thread(ve1); //t1 and t2 operate on same instance of VolatileExample class  

     t1.setName("t1");
     t2.setName("t2");

     t1.start();
     t2.start();
 }
}

输出:

Inside run when vol is 5. Thread is : t1
Inside run when vol is 5. Thread is : t1
Inside run when vol is 5. Thread is : t2
Inside run when vol is 5. Thread is : t1
Got Thread : t2 Now Calling Stop To End This Flow
Inside run when vol is 5. Thread is : t1

对 vol 变量的所有写入都将立即写入主内存，并且应该立即对其他线程“可见”。为什么调用 stopIt() 后 t1 线程仍然执行？难道它看不到 vol 值现在是 10 而不是 5 吗？

Answer 1

没有证据表明 t1 在调用 stopIt() 后运行。

事情可能按这个顺序发生

     t1                     t2
                         System.out.println("calling stopIt()");
while(vol==5)
System.out.println("Inside run")
                         enter stopIt()
                         vol = 10

这会给你你观察到的结果。 (还有其他可能的排序方式可以给您带来这个结果。)