java多线程-竞态条件的形成模拟，锁的使用，volatile关键字的使用

很简单的一次尝试。
售票的机器类public class TicketSell implements Runnable{
private volatile int number = 10; //第四次尝试去掉了volatile

@Override
public void run() {
while (number > 0) {
sold();
}
}

public synchronized void sold(){ //只有第三次的时候加了synchronized
if (number > 0) {
System.out.println(Thread.currentThread().getName()+"sold "+number);
number--;
}
}

}售票主函数public class TicketMain {
public static void main(String[] args) {
TicketSell thread = new TicketSell();
TicketSell thread2 = new TicketSell();
Thread t1 = new Thread(thread);
Thread t2 = new Thread(thread);
t1.setName("t1");
t2.setName("t2");
t1.start();
t2.start();
}
}第一次：
使用thread和thread1对象 两个线程分别跑起来，把thread对象和thread1对象传进去，相当于两个机器分别卖各自的票，各自有10张票
运行结果：t2 sold 10
t2 sold 9
t1 sold 10
t2 sold 8
t1 sold 9
t2 sold 7
t1 sold 8
t2 sold 6
t1 sold 7
t2 sold 5
t1 sold 6
t1 sold 5
t1 sold 4
t1 sold 3
t1 sold 2
t1 sold 1
t2 sold 4
t2 sold 3
t2 sold 2
t2 sold 1

Process finished with exit code 0互相不受影响，按照10-1顺序减下来
第二次：
然后将thread都传进t1和t2，类似于一个售票机分身两个（这样好理解？）然后类似于同时工作
运行结果是这样的（可能还有很神奇的负数或者0出现，因为出现了竞态条件）t2 sold 10
t1 sold 10
t2 sold 9
t1 sold 8
t2 sold 7
t1 sold 6
t2 sold 5
t2 sold 3
t1 sold 4
t2 sold 2
t1 sold 1

Process finished with exit code 0第三次：
加上了synchronized之后，就实现了正常的效果t1 sold 10
t1 sold 9
t1 sold 8
t1 sold 7
t1 sold 6
t2 sold 5
t2 sold 4
t2 sold 3
t2 sold 2
t2 sold 1

Process finished with exit code 0第四次：

volatile作用：用于状态的存放，共享，确保变量的更新操作通知到其他线程。

前面几次都是在volatile变量使得可见性得到确保  然后不会出现这样的情况t2 sold 10
t1 sold 10
t2 sold 9
t2 sold 7
t2 sold 6
t2 sold 5
t2 sold 4
t2 sold 3
t2 sold 2
t2 sold 1
t1 sold 8

Process finished with exit code 0t1最后还在卖8号票 说明可见性出现了问题。