1.11 Synchronized在遇到异常时, 默认情况下,

/**
 * This is description.
 * Synchronized在遇到异常时, 默认情况下, 会释放锁.
 *
 * @author Chris Lee
 * @date 2019/3/4 22:52
 */
public class Demo implements Runnable {

    private int count = 0;
    boolean flag = true;

    @Override
    public synchronized void run() {
        while (flag) {
            count++;
            System.out.println(Thread.currentThread().getName() + ", count: " + count);
            try {
                TimeUnit.SECONDS.sleep(1);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            if (count == 5) {
                int i = 1 / 0;
                System.out.println(i);
            }
        }
    }

    public static void main(String[] args) {
        Demo demo = new Demo();

        new Thread(() -> demo.run(), "t1").start();
        new Thread(() -> demo.run(), "t2").start();
        /*
            t1, count: 1
            t1, count: 2
            t1, count: 3
            t1, count: 4
            t1, count: 5
            t2, count: 6
            Exception in thread "t1" java.lang.ArithmeticException: / by zero
                at com.chris.concurrency1.c011synmeetexception.Demo.run(Demo.java:28)
                at com.chris.concurrency1.c011synmeetexception.Demo.lambda$main$0(Demo.java:37)
                at java.lang.Thread.run(Thread.java:745)
            t2, count: 7
            t2, count: 8
            t2, count: 9
            t2, count: 10
         */
    }
}

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资料:

1. 学习视频: https://www.bilibili.com/video/av11076511/?p=1
2. 参考代码: https://github.com/EduMoral/edu/tree/master/concurrent/src/yxxy
3. 我的代码: https://github.com/ChrisLeejing/learn_concurrency.git