使用java中的两个线程多次打印"Hello"和"world"

Question

Assume that one thread prints "Hello" and another prints "World". I have done it successfully for one time, as follows: package threading;

public class InterThread {

    public static void main(String[] args) {
        MyThread mt=new MyThread();
        mt.start();
        synchronized(mt){
            System.out.println("Hello");
            try {
                mt.wait();
                i++;
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }

    }

}

class MyThread extends Thread{

    public void run(){
        synchronized(this){
        System.out.println("World!");
        notify();
        }
    }
}

How do I do it for multiple time printing, say for 5 times? I tried putting for loop around the synchronized block, but of no use.

Answer 1

这是两个相互依赖的线程,我们需要两个同步对象.它们可能是众多事物之一.一个整数,另一个对象; 一个布尔另一个对象; 两个对象; 两个信号量等等.同步技术可以是你喜欢的Monitor或Semaphore,但它们必须是两个.

我已修改您的代码以使用信号量而不是Monitor.信号量工作更透明.您可以看到获取和发布的情况.监视器甚至是更高的构造.因此同步在引擎盖下工作.

如果您对以下代码感到满意,则可以将其转换为使用监视器.

    import java.util.concurrent.Semaphore;

    public class MainClass {

        static Semaphore hello = new Semaphore(1);
        static Semaphore world = new Semaphore(0);

        public static void main(String[] args) throws InterruptedException {
            MyThread mt=new MyThread();     
            mt.hello = hello;
            mt.world = world;
            mt.start();

            for (int i=0; i<5; i++) {
                hello.acquire(); //wait for it
                System.out.println("Hello");

                world.release(); //go say world
            }
        }
    }

    class MyThread extends Thread{

        Semaphore hello, world;

        public void run(){
            try {
                for(int i = 0; i<5; i++) {
                    world.acquire(); // wait-for it
                    System.out.println("  World!");

                    hello.release(); // go say hello
                }

            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }