synchronized如何实现两个线程交替运行？看完你就懂了，列害dei

Posted 2020-12-20

背景

用两个线程交替输出A-Z和1-26，即一个线程输出A-Z，另一个线程输出1-26
而且是交替形式

线程1输出A——线程二输出1
线程1输出B——线程二输出2
线程1输出C——线程二输出3
以此类推

分析

主要考察线程之间的通信，思路就是创建两个线程

在一个线程输出一个内容之后，自己进入阻塞，去唤醒另一个线程

另一个线程同样，输出一个内容之后，自己进入阻塞，去唤醒另一个线程

代码实现（一）

public class AlternateCover {

    public static void main(String[] args) {

        final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
        final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};

        threadRun(arrLetter, arrNumber);
    }

    private static void threadRun(char[] arrLetter,String[] arrNumber){

        final Object lock = new Object();// 设置一个锁对象

        // print arrNumber
        new Thread(() -> {
            synchronized (lock) {
                for (String a : arrNumber) {
                    System.out.print( a);
                    try {
                        lock.notify();// 唤醒其他等待的线程 此处唤醒 arrLetter
                        lock.wait();// arrNumber自己进入等待 让出CPU资源和锁资源
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();
            }
        }, "arrNumber ").start();

        // print arrLetter
        new Thread(() -> {
            synchronized (lock) {// 获取对象锁
                for (char a : arrLetter) {
                    System.out.print(a);
                    try {
                        lock.notify();// 唤醒其他等待的线程 此处唤醒 arrNumber
                        lock.wait();// arrLetter自己进入等待 让出CPU资源和锁资源
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
            }
        }, "arrLetter ").start();

    }
}

运行一下，确实实现了交替输出，但是多运行几次，就会发现问题

有时候是数字先输出，有时候是字母先输出

即两个线程谁先启动的顺序是不固定的

倘若试题中再加一句，必须要字母先输出，怎么办？

代码实现（二）

/**
 * 交替掩护 必须保证大写字母先输出
 */
public class AlternateCover {

    public static volatile Boolean  flg = false;// 谁先开始的标志 volatile修饰目的是让该值修改对所有线程可见，且防止指令重排序
    public static void main(String[] args) {

        final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
        final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};

        threadRun(arrLetter, arrNumber);
    }

    private static void threadRun(char[] arrLetter,String[] arrNumber){

        final Object lock = new Object();// 锁对象

        // print arrLetter
        new Thread(() -> {
            synchronized (lock) {
                if (!flg){ // 如果flg是false 就将值设为true
                    flg = true;
                }
                for (char a : arrLetter) {
                    System.out.print(a);// 输出内容
                    try {
                        lock.notify();// 唤醒在等待的其他线程中的一个（此处也只有另一个）
                        lock.wait();// 自己进入等待 让出CPU资源和锁资源
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
            }
        }, "arrLetter").start();

        // print arrNumber
        new Thread(() -> {
            synchronized (lock) {
                if (!flg){// 倘若是该线程先执行，那么flg次数还是false 就先等着
                    try {
                        lock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }

                for (String a : arrNumber) {
                    System.out.print( a);
                    try {
                        lock.notify();
                        lock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();
            }
        }, "arrNumber").start();

    }

}

如此问题可以得到解决，但有更优(装)雅（B）的解决办法

CountDownLatch实现

/**
 * 交替掩护 必须保证大写字母先输出
 */
public class AlternateCover {

    private static CountDownLatch count = new CountDownLatch(1);// 计数器容量为1
    public static void main(String[] args) {

        final char[] arrLetter = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
        final String[] arrNumber = {"1","2","3","4","5","6","7","8","9","10","11","12","13","14","15","16","17","18","19","20","21","22","23","24","25","26"};

        threadRun(arrLetter, arrNumber);
    }

    private static void threadRun(char[] arrLetter,String[] arrNumber){

        final Object lock = new Object();

        // print arrLetter
        new Thread(() -> {
            synchronized (lock) {// 获取对象锁
                count.countDown();// 对计数器进行递减1操作，当计数器递减至0时，当前线程会去唤醒阻塞队列里的所有线程（只针对count）
                for (char a : arrLetter) {
                    System.out.print(a);
                    try {
                        lock.notify();// 唤醒其他等待的线程 此处唤醒 arrNumber
                        lock.wait();// arrLetter自己进入等待 让出CPU资源和锁资源
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();// 最后那个等待的线程需要被唤醒，否则程序无法结束
            }
        }, "arrLetter ").start();

        // print arrNumber
        new Thread(() -> {
            synchronized (lock) {
                try {
                    count.await();// 如果该线程先执行 阻塞当前线程，将当前线程加入阻塞队列
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                for (String a : arrNumber) {
                    System.out.print( a);
                    try {
                        lock.notify();// 唤醒其他等待的线程 此处唤醒 arrLetter
                        lock.wait();// arrNumber自己进入等待 让出CPU资源和锁资源
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }
                lock.notify();
            }
        }, "arrNumber ").start();

    }

}

结尾

以上就是synchronized实现两个线程交替运行的解法，欢迎留言交流
感谢看到最后的朋友，都看到最后了，点个赞再走啊，如有不对之处还请多多指正。