编写一个带有2个线程的程序,可以交替打印

Question

最近我在接受采访时被问到了这个问题.

编写一个带有两个线程(A和B)的程序,其中A打印1,B打印2,依此类推,直到达到50.

我们该怎么做呢？

Answer 1

赋值的本质是演示线程如何发出另一个信号.最常见的方法是使用阻塞队列,但此处信号不携带任何信息,因此信号量就足够了.

创建用2个信号量参数化的线程类:输入和输出:

class ThreadPrinter implements Runnable {
    int counter;
    Semaphore ins, outs;

    ThreadPrinter(int counter, Semaphore ins, Semaphore outs) {
        this.counter = counter;
        this.ins = ins;
        this.outs = outs;
    }

    @Override
    public void run() {
        for (int i = 0; i < 25; i++) {
            ins.aquire(); // wait for permission to run
            System.out.println("" + counter);
            outs.release();  // allow another thread to run
            counter += 2;
        }
    }

创建2 Semaphore秒并将它们传递给2个线程:

Semaphore a = new Semaphore(1);  // first thread is allowed to run immediately
Semaphore b = new Semaphore(0); // second thread has to wait
ThreadPrinter tp1 = new ThreadPrinter(1, a, b);
ThreadPrinter tp2 = new ThreadPrinter(2, b, a); 

注意信号量a并b以不同的顺序传递.

Answer 2

这是另一个解决方案：

     Thread t1 = new Thread(new Runnable() {

        @Override
        public void run() {
            synchronized (lock) {
                for (int i = 1; i <= 50; i += 2) {
                    System.out.println("T1=" + i);

                    t1turn = false;
                        try {
                            lock.notifyAll();
                            lock.wait();
                        } catch (InterruptedException e) {
                        }
                }
            }

        }
    });
    Thread t2 = new Thread(new Runnable() {

        @Override
        public void run() {
            synchronized (lock) {
                for (int i = 2; i <= 50; i += 2) {
                    if (t1turn)
                        try {
                            lock.wait();
                        } catch (InterruptedException e) {
                        }
                    System.out.println("T2=" + i);
                    t1turn = true;
                    lock.notify();
                }
            }
        }
    });
    t1.start();
    t2.start();