【多线程例题】顺序打印abc线程

顺序打印-进阶版

方法一：三个线程竞争同一个锁，通过count判断是否打印

方法二：三个线程同时start，分别上锁，从a开始，打印后唤醒b

三个线程分别打印A，B，C
方法一：通过count计数打印（三个线程上同样的锁，打印一个，召唤所有锁，如果不满足条件，则wait等待，锁自动解锁）
方法二：

/*** 有三个线程，分别只能打印A，B和C* 要求按顺序打印ABC，打印10次* 输出示例：* ABC* ABC* ABC* ABC* ABC* ABC* ABC* ABC* ABC* ABC*/
public class Thread_ {// 计数器private static volatile int COUNTER = 0;// 定义一个单独的锁对象private static Object lock = new Object();public static void main(String[] args) {// 创建三个线程，并指定线程名，每个线程名分别用A,B,C表示Thread t1 = new Thread(() -> {// 循环10次for (int i = 0; i < 10; i++) {// 执行的代码加锁synchronized (lock) {// 每次唤醒后都重新判断是否满足条件// 每条线程判断的条件不一样，注意线程t1,t2while (COUNTER % 3 != 0) {try {// 不满足输出条件时，主动等待并释放锁lock.wait();} catch (InterruptedException e) {e.printStackTrace();}}// 满足输出条件，打印线程名，每条线程打印的内容不同System.out.print(Thread.currentThread().getName());// 累加计数COUNTER++;// 唤醒其他线程lock.notifyAll();}}}, "A");Thread t2 = new Thread(() -> {for (int i = 0; i < 10; i++) {synchronized (lock) {while (COUNTER % 3 != 1) {try {lock.wait();} catch (InterruptedException e) {e.printStackTrace();}}System.out.print(Thread.currentThread().getName());COUNTER++;lock.notifyAll();}}}, "B");Thread t3 = new Thread(() -> {for (int i = 0; i < 10; i++) {synchronized (lock) {while (COUNTER % 3 != 2) {try {lock.wait();} catch (InterruptedException e) {e.printStackTrace();}}// 换行打印System.out.println(Thread.currentThread().getName());COUNTER++;lock.notifyAll();}}}, "C");// 启动线程t1.start();t2.start();t3.start();}
}

public class Demo {private static Object locker1 = new Object();private static Object locker2 = new Object();private static Object locker3 = new Object();public static void main(String[] args) throws InterruptedException {Thread t1 = new Thread(() -> {try {for (int i = 0; i < 10; i++) {synchronized (locker1) {locker1.wait();}System.out.print("A");synchronized (locker2) {locker2.notify();}}} catch (InterruptedException e) {e.printStackTrace();}});Thread t2 = new Thread(() -> {try {for (int i = 0; i < 10; i++) {synchronized (locker2) {locker2.wait();}System.out.print("B");synchronized (locker3) {locker3.notify();}}} catch (InterruptedException e) {e.printStackTrace();}});Thread t3 = new Thread(() -> {try {for (int i = 0; i < 10; i++) {synchronized (locker3) {locker3.wait();}System.out.println("C");synchronized (locker1) {locker1.notify();}}} catch (InterruptedException e) {e.printStackTrace();}});t1.start();t2.start();t3.start();Thread.sleep(1000);// 从线程 t1 启动synchronized (locker1) {locker1.notify();}}
}