Java synchronized线程交替运行实现过程详解

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();

  }

}

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