编写一个程序,开启3个线程,这三个线程的ID分别为A、B、C,每个线程将自己的ID在屏幕上打印10遍,要求输出的结果必须按顺序显示。

编写一个程序,开启3个线程,这三个线程的ID分别为A、B、C,每个线程将自己的ID在屏幕上打印10遍,要求输出的结果必须按顺序显示。

一道Java并发相关的面试题。本文使用Lock+Condition实现。

package com.xudc.juc;

import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

/**
 * 编写一个程序,开启3个线程,这三个线程的ID分别为A、B、C,每个线程将自己的ID在屏幕上打印10遍,要求输出的结果必须按顺序显示。
 * 如:ABCABCABCABC...依次递归
 * @author xudc
 */
public class TestAlternate {
    public static void main(String[] args) {
        Alternate alternate = new Alternate();
        final int count = 10;
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printA();
            }
        },"A").start();
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printB();
            }
        },"B").start();
        new Thread(() -> {
            for (int i = 0; i < count; i++) {
                alternate.printC();
            }
        },"C").start();
    }
}

class Alternate{
    private final Lock lock = new ReentrantLock();
    private Condition conditionA = lock.newCondition();
    private Condition conditionB = lock.newCondition();
    private Condition conditionC = lock.newCondition();

    private int number = 1;

    public void printA(){
        lock.lock();
        try {
            while (number != 1) {
                conditionA.await();
            }
            System.out.print("A");
            number = 2;
            conditionB.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }

    public void printB(){
        lock.lock();
        try {
            while (number != 2) {
                conditionB.await();
            }
            System.out.print("B");
            number = 3;
            conditionC.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }

    public void printC(){
        lock.lock();
        try {
            while (number != 3) {
                conditionC.await();
            }
            System.out.print("C");
            number = 1;
            conditionA.signal();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            lock.unlock();
        }
    }
}

运行结果:

ABCABCABCABCABCABCABCABCABCABC

原文地址:https://xudc.tech/2019/06/01/thread-interview-1/

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转载自blog.csdn.net/E09620126/article/details/90733684