No4.传统线程同步通信技术

1.面试题

   子线程循环10次，接着住线程循环100次，接着又回到子线程循环10次，接着再回到主线程又循环100次，如此循环50次。

 

public class ThreadTest {

    public static void main(String[] args) {
        new ThreadTest().init();

    }

    public void init() {
        final Business business = new Business();
        new Thread(new Runnable() {

            public void run() {
                for (int i = 0; i < 50; i++) {
                    business.SubThread(i);
                }
            }

        }

        ).start();

        for (int i = 0; i < 50; i++) {
            business.MainThread(i);
        }
    }

    private class Business {
        boolean bShouldSub = true;// 这里相当于定义了控制该谁执行的一个信号灯

        public synchronized void MainThread(int i) {
            if (bShouldSub)
                try {
                    this.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }

            for (int j = 0; j < 100; j++) {
                System.out.println(Thread.currentThread().getName() + ":i=" + i
                        + ",j=" + j);
            }
            bShouldSub = true;
            this.notify();

        }

        public synchronized void SubThread(int i) {
            if (!bShouldSub)
                try {
                    this.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }

            for (int j = 0; j < 10; j++) {
                System.out.println(Thread.currentThread().getName() + ":i=" + i
                        + ",j=" + j);
            }
            bShouldSub = false;
            this.notify();
        }

    }

}