java - 两个线程进入 runnable 的同步方法

Question

这是我遇到问题的代码 --

public class WaitTest {
    public static void main(String[] args) {
        Runner rr = new Runner();
        Thread t1 = new Thread(rr,"T1");
        Thread t2 = new Thread(rr,"T2");
        t1.start();
        t2.start();
    }
}

class Runner implements Runnable{
    int i=0;
    public void run(){
        try{
            if(Thread.currentThread().getName().equals("T1")){
                bMethod();
                aMethod();
            }
            else{
                aMethod();
                bMethod();
            }
        }catch(Exception e){}
    }

    public synchronized void aMethod() throws Exception{
        System.out.println("i="+i+",Now going to Wait in aMethod "+Thread.currentThread().getName());
        Thread.currentThread().wait();
        System.out.println("Exiting aMethod "+Thread.currentThread().getName());
    }

    public synchronized void bMethod() throws Exception{
        System.out.println("i="+i+",Now going to Wait bMethod "+Thread.currentThread().getName());
        i=5;
        notifyAll();
        System.out.println("Exiting bMethod "+Thread.currentThread().getName());
    }
}

输出是:

i=0,Now going to Wait bMethod T1
Exiting bMethod T1
i=5,Now going to Wait in aMethod T1
i=5,Now going to Wait in aMethod T2

我的问题是:

Why T2 enters in aMethod while T1 is waiting inside? and Why T2 prints i=5 in aMethod.

Answer 1

当你执行wait时，你的线程释放锁并进入等待状态。这时允许其他线程进入该方法。您正在使用 Runnable 的单个实例，因此当一个线程将其设置为 5 时，另一个线程将读取该值。