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还是一个的线程问题

msl12 发布于 2015-10-03 15:58, 574 次点击
代码如下:(问题在运行结果那里)
-----------------------------------------------------------------------------------------------------------------------------------
class MyThread implements Runnable {
    private int ticket = 5;

    public void run() {
        for (int i = 0; i < 5; ++i) {
            //synchronized (this) {
                if (ticket > 0) {
                    try {
                        //Thread.sleep(500);
                    } catch (Exception e) {
                        e.printStackTrace();
                    }

                    System.out.println(Thread.currentThread().getName() + "卖票: ticket = " + (ticket--));
                }
            //}
        }
    }
}

public class Test2 {
    public static void main(String[] agrs) {
        MyThread mt = new MyThread();
        Thread t1 = new Thread(mt);
        Thread t2 = new Thread(mt);
        Thread t3 = new Thread(mt);

        t1.start();
        t2.start();
        t3.start();
    }
}

一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一
运行结果却是:
------------------------------------------------------------------------------------------------------------------------------------------
Thread-1卖票: ticket = 3
Thread-2卖票: ticket = 5                //怎么5会在3的下面?
Thread-0卖票: ticket = 4
Thread-0卖票: ticket = 2
Thread-1卖票: ticket = 2
Thread-2卖票: ticket = 2
Thread-2卖票: ticket = 1
Thread-1卖票: ticket = 1
Thread-0卖票: ticket = 0
4 回复
#2
林月儿2015-10-03 17:49
以下是引用msl12在2015-10-3 15:58:33的发言:

代码如下:(问题在运行结果那里)
-----------------------------------------------------------------------------------------------------------------------------------
class MyThread implements Runnable {
    private int ticket = 5;

    public void run() {
        for (int i = 0; i < 5; ++i) {
            //synchronized (this) {
                if (ticket > 0) {
                    try {
                        //Thread.sleep(500);
                    } catch (Exception e) {
                        e.printStackTrace();
                    }

                    System.out.println(Thread.currentThread().getName() + "卖票: ticket = " + (ticket--));
                }
            //}
        }
    }
}

public class Test2 {
    public static void main(String[] agrs) {
        MyThread mt = new MyThread();
        Thread t1 = new Thread(mt);
        Thread t2 = new Thread(mt);
        Thread t3 = new Thread(mt);

        t1.start();
        t2.start();
        t3.start();
    }
}

一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一一
运行结果却是:
------------------------------------------------------------------------------------------------------------------------------------------
Thread-1卖票: ticket = 3
Thread-2卖票: ticket = 5                //怎么5会在3的下面?
Thread-0卖票: ticket = 4
Thread-0卖票: ticket = 2
Thread-1卖票: ticket = 2
Thread-2卖票: ticket = 2
Thread-2卖票: ticket = 1
Thread-1卖票: ticket = 1
Thread-0卖票: ticket = 0


加个锁不就行啦,多线程访问同一资源事务一致性问题需要解决
#3
msl122015-10-03 18:56
回复 2楼 林月儿
就是奇怪ticket不应该是从5才能到3么,第一行和第二行却显示的是奇怪的内容?
线程这里难理解...
#4
林月儿2015-10-03 20:52
以下是引用msl12在2015-10-3 18:56:58的发言:

就是奇怪ticket不应该是从5才能到3么,第一行和第二行却显示的是奇怪的内容?
线程这里难理解...


都说加个锁就好了,没啥难理解的
#5
神vLinux飘飘2015-10-06 23:10
5在3下边不奇怪,主要原因是

当这段代码被执行的时候
程序代码:

t1.start();
t2.start();
t3.start();

你必须明白,t1/t2/t2中的run()方法未必按照你代码执行的顺序而执行(取决于OS的调度进程),甚至t1/t2/t3执行的顺序jvm都不做任何保证(重排序)。


我们来给你的输出按照线程编号排个顺序,恩,至少对每个线程而言,ticket是从大减到小的 :)

程序代码:

Thread-0卖票: ticket = 4
Thread-0卖票: ticket = 2
Thread-0卖票: ticket = 0


程序代码:

Thread-1卖票: ticket = 3
Thread-1卖票: ticket = 2
Thread-1卖票: ticket = 1


程序代码:

Thread-2卖票: ticket = 5
Thread-2卖票: ticket = 2
Thread-2卖票: ticket = 1


恩,这样看来,似乎5在3前边也很好理解的。
至于为什么将synchronized(this)的注释打开就好了的问题,再之前的帖子中已经给你大概提了下,synchronized(this)锁的就是mt对象


MyThread mt = new MyThread();


当注释打开后,t1/t2/t3三个线程在执行到synchronized(this)时都必须等待持有mv锁的线程释放之后才能继续执行,所以你的票不会超卖。

[ 本帖最后由 神vLinux飘飘 于 2015-10-6 23:16 编辑 ]
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