[英]Java Thread not go correctly
我想改变Thread
机制,所以我做了一个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");
}
}
运行该应用程序时,出现了意外的结果:
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
但是在另一种实践中, 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
这些使我感到困惑:
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。 ThreadTest
添加Thread.yield()或Thread.sleep(),t1 / t2 /主线程可以单独运行。 我如何修改我的Company
代码以使其符合我的期望(问题1〜3),为什么?
非常感谢。
唯一的问题是endCount
上没有同步,因此当Tom
和Mary
调用Boss
onEnd
方法并递增endCount
,工作的Boss
可能不会注意到它。
您可以使用AtomicInteger
,
AtomicInteger endCount
和
endCount.incrementAndGet()
代替endCount++
endCount.get() < members.size()
代替endCount<members.size()
这样JMM可以保证Boss
在其循环中获得新的价值。
并且,正如注释中所建议的,您可以在Worker
循环中添加此代码,这将更容易模拟multi-thread
环境:
try {
Thread.sleep(10);
} catch (Exception e) {
}
更新
启动多个线程时,如果不同步,则无法控制它们中每一行的执行顺序。 它们的执行顺序由CPU安排。 您可以检查一下 。
即使在第二次测试中,除了Main: Start!
确保在第一行显示,其他行的顺序仍然不确定 。
而且, Thread.sleep
或Thread.yield
只会使仿真并发执行变得更加容易,但仍不保证Tom
和Mary
会在控制台上逐行输出内容。
这是我计算机上的测试结果:
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!
我希望汤姆和玛丽可以一起工作,但结果是玛丽在汤姆工作结束后开始工作
因为您没有让线程产生任何逻辑位置(例如IO),所以不能保证上下文会切换。 调度程序不会在此处强制执行此操作,因此它们都可以运行直到串行完成为止。
您可以通过添加Thread.yield()
(或进行睡眠或执行一些IO)来强制执行此操作。
即使效果差不多, sleep
和yield
意味着不同的事情。
sleep
说:“亲爱的调度程序,此线程不希望在接下来的x毫秒内执行任何操作” yield
说:“亲爱的调度程序现在是让其他线程做事情的好时机”。 选择最接近您想要说的一种。
我希望主线程应该在每个IWork对象开始工作后结束,但似乎没有
与上述相同。 您所有的线程都是繁忙的循环,占用CPU直到完成。
终于,老板似乎永不停止工作。
根据下面的评论,endCount不是线程安全的。 您可以将其包装在AtomInteger中,或添加一些同步块。
顺带一提,您应该考虑Boss根本不是线程。 她可以仅通过回调来实现。
[更新/附加此问题:]我不必为ThreadTest添加Thread.yield()或Thread.sleep(),t1 / t2 /主线程可以单独运行。
调度程序按照调度程序的要求进行操作。 如果您第一次运行多次,则不能保证在任何一个应用程序中都能获得相同的结果。
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