java wait超时_解决Java的wait(long mills)方法不能区分其返回是由于超时还是被唤醒的问题...

wait(long mills) 没有返回值，所以区分不了其返回是由于超时还是被唤醒，因此需要引入一个布尔变量，来表示它的返回类型。

class WaitTimeOut {

private volatile boolean ready = false; // 如果是true，则表示是被唤醒

public synchronized void notify0() {

ready = true;

notify();

}

public synchronized void wait0(long mills) throws InterruptedException {

long begin = System.currentTimeMillis();

long rest = mills;

if (rest == 0L) {

wait(0);

} else {

while (!ready && rest > 0) { // 如果被唤醒(ready为true)，或超时(rest <= 0)则结束循环

wait(rest);

rest = mills - (System.currentTimeMillis() - begin); // 计算剩余时间

}

if (ready) {

System.out.println("被唤醒");

} else {

System.out.println("超时退出");

}

}

}

public static void main(String[] args) {

ExecutorService pool = Executors.newFixedThreadPool(2);

WaitTimeOut waiter = new WaitTimeOut();

pool.execute(() -> {

try {

waiter.wait0(10000);

} catch (InterruptedException e) {

e.printStackTrace();

}

});

pool.execute(waiter::notify0);

}

}