Java 多线程售票

方式一：继承thread 类

每个线程100张票 

package com.example.starter;

public class SellTickets extends Thread{

  public static void main(String[] args) {
    Thread t1 = new SellTick1();
    Thread t2 = new SellTick1();
    Thread t3 = new SellTick1();
    t1.start();
    t2.start();
    t3.start();
  }
}

class SellTick1 extends Thread{
  private int num = 100;

  @Override
  public void run(){
      while(true){
        if(num <= 0){
          System.out.println("售票结束");
          break;
        }
        try {
          Thread.sleep(50);
        } catch (InterruptedException e) {
          e.printStackTrace();
        }
        System.out.println("窗口"+Thread.currentThread().getName()+ "售出 一张票 剩余票数=" + (--num ));
      }
  }
}

结果： 

 方式二：num 使用static修饰 

一共100张票 有重复减票的情况，最后还有减成负数的情况

package com.example.starter;

public class SellTickets extends Thread{

  public static void main(String[] args) {
    Thread t1 = new SellTick1();
    Thread t2 = new SellTick1();
    Thread t3 = new SellTick1();
    t1.start();
    t2.start();
    t3.start();
  }
}

class SellTick1 extends Thread{
  private static int num = 100;

  @Override
  public void run(){
      while(true){
        if(num <= 0){
          System.out.println("售票结束");
          break;
        }
        try {
          Thread.sleep(50);
        } catch (InterruptedException e) {
          e.printStackTrace();
        }
        System.out.println("窗口"+Thread.currentThread().getName()+ "售出 一张票 剩余票数=" + (--num));
      }
  }
}

结果：

方式三：实现runnable的方式

package com.example.starter;

public class SellTickets extends Thread{

  public static void main(String[] args) {
    SellTicket2 sellTicket2 = new SellTicket2();
    Thread t1 = new Thread(sellTicket2);
    Thread t2 = new Thread(sellTicket2);
    Thread t3 = new Thread(sellTicket2);
    t1.start();
    t2.start();
    t3.start();
  }
}

class SellTicket2 implements   Runnable{
  private  int num = 100;
  @Override
  public void run() {
    while(true){
      if(num <= 0){
        System.out.println("售票结束");
        break;
      }
      try {
        Thread.sleep(50);
      } catch (InterruptedException e) {
        e.printStackTrace();
      }
      num --;
      System.out.println("窗口"+Thread.currentThread().getName()+ "售出 一张票 剩余票数=" + (num));
    }
  }
}

结果

方式四：以上三种方式多存在多售票的问题，使用synchronized优化

package com.example.starter;

public class SellTickets extends Thread{

  public static void main(String[] args) {
    SellTicket2 sellTicket2 = new SellTicket2();
    Thread t1 = new Thread(sellTicket2);
    Thread t2 = new Thread(sellTicket2);
    Thread t3 = new Thread(sellTicket2);
    t1.start();
    t2.start();
    t3.start();
  }
}

class SellTicket2 implements   Runnable{
  private volatile int num = 100;
  private volatile  boolean loop = true;
  @Override
  public  void run() {
    while(loop){
      sell();
    }
  }

  synchronized void  sell(){
    if(num <= 0){
      System.out.println("售票结束");
      loop = false;
      return;
    }
    num --;
    System.out.println("窗口"+Thread.currentThread().getName()+ "售出 一张票 剩余票数=" + (num));
  }
}

结果：

方式五：

package com.example.starter;

public class SynSellTickets extends Thread{

  public static void main(String[] args) {
    SellTicket3 sellTicket3 = new SellTicket3();
    Thread t1 = new Thread(sellTicket3);
    Thread t2 = new Thread(sellTicket3);
    Thread t3 = new Thread(sellTicket3);
    t1.start();
    t2.start();
    t3.start();
  }
}

class SellTicket3 implements   Runnable{
  private  int num = 100;
  private  boolean loop = true;
  @Override
  public  void run() {
    while(loop){
      synchronized (this){
        if(num > 0){
          num --;
          System.out.println("窗口"+Thread.currentThread().getName()+ "售出 一张票 剩余票数=" + (num));
        }else{
          loop = false;
        }
      }
    }
    System.out.println("售票结束");
  }

}

结果：