多线程Java线程运行不正常

1 周，3 日 Questions & Answers 142

我想改变Thread机制，所以我创建了一个Company类，如下所示：

public class Company {
    static void p(String s){ System.out.println(s); }

    interface IWork{ void work(); }
    interface OnReportListener{ int onEnd(Worker w); }
    static class Job{ int effertCount, budget=100; }

    static class Worker implements IWork{
        String name; Job job=new Job(); OnReportListener listener; boolean isOver;
        public Worker(String n, OnReportListener l) {
            name = n; listener = l;
        }

        public void work() {
            new Thread(){
                public void run() {
                    while (!isOver) {
                        int spent = (int) Math.round(Math.random()*7-2) ;
                        if (spent<0) p(name+": I earned $"+(-spent));
                        isOver = (job.budget-=spent) <=0;
                        job.effertCount++;
                    }
                    p(name+": OMG, I got the salary $"+ listener.onEnd(Worker.this));
                }
            }.start();
        }
    }

    static class Boss implements IWork, OnReportListener{
        Set<Worker> members; int endCount;
        public Boss(Set<Worker> s){ members = s;}
        public int onEnd(Worker w) {
            p("Boss: "+w.name+", thanks for your effort, you deserve it!");
            endCount++;
            return w.job.effertCount*10;
        }

        public void work() {
            new Thread(){
                public void run() {
                    while (endCount<members.size()) { /*fool around*/ }
                    p("Boss: It's time to go home!");
                }
            }.start();
        }
    }

    public static void main(String[] args) {
        Set<Worker> workers = new HashSet<Worker>();
        Boss boss = new Boss(workers);
        Worker tom = new Worker("Tom", boss); 
        workers.add(tom); // hire Tom
        Worker mary = new Worker("Mary", boss); 
        workers.add(mary); // hire Mary

        p("Company.main: Start to work!");
        boss.work();
        tom.work();
        mary.work();
        p("Company.main: End of the assigning");
    }
}

当我运行应用程序时，我得到了意想不到的结果：

Company.main: Start to work!
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $2
Tom: I earned $2
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $2
Tom: I earned $2
Tom: I earned $1
Tom: I earned $1
Tom: I earned $1
Tom: I earned $2
Tom: I earned $1
Boss: Tom, thanks for your effort, you deserve it!
Tom: OMG, I got the salary $770
Mary: I earned $1
Mary: I earned $2
Mary: I earned $2
Mary: I earned $1
Mary: I earned $1
Mary: I earned $1
Mary: I earned $2
Mary: I earned $1
Mary: I earned $2
Mary: I earned $1
Mary: I earned $1
Mary: I earned $2
Boss: Mary, thanks for your effort, you deserve it!
Mary: OMG, I got the salary $510
Company.main: End of the assigning

但在另一种实践中，ThreadTest类：

public class ThreadTest extends Thread{
    static void p(String s){ System.out.println(s); }
    public ThreadTest(String s){ super(s); }
    public void run() {
        for (int i = 0; i < 25; i++) p(getName()+": "+i);
    }

    public static void main(String[] args) {
        p("Main: Start!");
        new ThreadTest("t1").start();
        new ThreadTest("t2").start();
        p("Main: Finish!");
    }
}

我运行它并得到：

Main: Start!
t1: 0
t1: 1
t1: 2
Main: Finish!
t2: 0
t2: 1
t2: 2
t2: 3
t2: 4
t2: 5
t1: 3
t1: 4
t1: 5
t1: 6
t2: 6
t2: 7
t2: 8
t2: 9
t2: 10
t2: 11
t2: 12
t2: 13
t2: 14
t2: 15
t2: 16
t2: 17
t2: 18
t2: 19
t2: 20
t2: 21
t2: 22
t2: 23
t2: 24
t1: 7
t1: 8
t1: 9
t1: 10
t1: 11
t1: 12
t1: 13
t1: 14
t1: 15
t1: 16
t1: 17
t1: 18
t1: 19
t1: 20
t1: 21
t1: 22
t1: 23
t1: 24

这些让我困惑：

我希望Company类的主线程应该在每个IWork对象开始工作后结束，但似乎没有
我希望汤姆和玛丽一起工作，但结果是玛丽在汤姆工作结束后工作
老板似乎从不停止他的工作
[更新/附加此问题：]我不必添加线程。屈服（）或线程。ThreadTest和t1/t2/主线程的sleep（）可以分别运行

我如何修改我的Company代码，让它符合我的期望（问题1~3），为什么

非常感谢

Tags:

共 (2) 个答案

# 1 楼答案
唯一的问题是endCount上没有同步，因此当Tom和Mary调用Boss的onEnd方法和增量endCount时，工作Boss可能不会注意到它

您可以使用AtomicInteger
```
AtomicInteger endCount
```
及
- 用endCount.incrementAndGet()代替endCount++
- 改用endCount.get() < members.size() endCount<members.size()
因此JMM可以保证Boss在其循环中获得新值

并且，正如注释中所建议的，您可以将其添加到Worker的循环中，这将更容易模拟multi-thread环境：
```
try {
    Thread.sleep(10);
} catch (Exception e) {

}
```
更新

当启动多个线程时，如果没有同步，就无法控制每行线程的执行顺序。它们的执行顺序由CPU安排。您可以选中this

即使在第二次测试中，除了保证在第一行显示Main: Start!之外，其他行的顺序仍然不确定

而且，Thread.sleep或Thread.yield只会使模拟并发执行变得更容易，它仍然不能保证Tom并且Mary会在控制台上逐行输出一些东西

这是我电脑上的测试结果：
```
Company.main: Start to work!
Company.main: End of the assigning
Tom: I earned $2
Mary: I earned $1
Mary: I earned $1
Mary: I earned $1
Tom: I earned $1
Tom: I earned $2
Mary: I earned $1
Tom: I earned $1
Tom: I earned $2
Mary: I earned $1
Mary: I earned $2
Tom: I earned $1
Mary: I earned $2
Mary: I earned $1
Mary: I earned $1
Mary: I earned $2
Mary: I earned $2
Mary: I earned $2
Tom: I earned $1
Tom: I earned $1
Mary: I earned $1
Tom: I earned $1
Mary: I earned $1
Tom: I earned $1
Mary: I earned $2
Tom: I earned $1
Mary: I earned $1
Tom: I earned $1
Mary: I earned $2
Mary: I earned $1
Tom: I earned $1
Boss: Tom, thanks for your effort, you deserve it!
Tom: OMG, I got the salary $590
Mary: I earned $1
Boss: Mary, thanks for your effort, you deserve it!
Mary: OMG, I got the salary $770
Boss: It's time to go home!
```
# 2 楼答案
I expect Tom and Mary should work together but the result is Mary works after the end of Tom's working

因为线程没有任何逻辑位置（例如IO），所以无法保证上下文会切换。调度程序在这里不强制执行，因此它们都以串行方式运行，直到完成

您可以通过添加Thread.yield()（或睡眠，或通过执行一些IO）来强制执行此操作

有了这个sleep和yield意味着不同的东西，尽管效果差不多
- sleep说“亲爱的调度程序，这个线程不想为下一个x毫秒做任何事情”
- yield说“亲爱的调度程序现在是让其他线程做事情的绝佳时机”
选择一个最接近你想说的话

I expect the main thread should end after each IWork objects starts to work but it seems doesn't

同上。所有线程都是繁忙的循环，占用CPU直到它们完成

Finally it seems the boss never stops his working...

根据下面的注释，endCount不是线程安全的。您可以将其包装在AtomInteger中，或者添加一些同步块

作为一个无关的人，你应该认为老板根本不是一根线。她可以通过回调来实现

[Update/Append this question:] I don't have to add Thread.yield() or Thread.sleep() for ThreadTest and t1/t2/main thread could run separately.

调度程序按照调度程序的要求执行。如果第一次运行多次，则不能保证在两个应用程序中都获得相同的结果

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多线程Java线程运行不正常

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