如何控制线程的执行顺序

错误示例：

 1 public class Test {
 2     public static void main(String[] args){
 3         thread1.start();
 4         thread2.start();
 5         thread3.start();
 6     }
 7 
 8     static Thread thread1 = new Thread(new Runnable() {
 9         @Override
10         public void run() {
11             System.out.println("thread1");
12         }
13     });
14 
15     static Thread thread2 = new Thread(new Runnable() {
16         @Override
17         public void run() {
18             System.out.println("thread2");
19         }
20     });
21 
22     static Thread thread3 = new Thread(new Runnable() {
23         @Override
24         public void run() {
25             System.out.println("thread3");
26         }
27     });
28 }

执行结果：

1 thread2
2 thread1
3 thread3

原因：start()方法只是让子线程处于就绪状态，最终执行状态是不可控的

方案一：

 1 public class Test {
 2     public static void main(String[] args) throws InterruptedException {
 3         thread1.start();
 4         thread1.join();//主线程放弃cpu执行权，让给子线程执行
 5         thread2.start();
 6         thread2.join();
 7         thread3.start();
 8         thread3.join();
 9     }
10 
11     static Thread thread1 = new Thread(new Runnable() {
12         @Override
13         public void run() {
14             System.out.println("thread1");
15         }
16     });
17 
18     static Thread thread2 = new Thread(new Runnable() {
19         @Override
20         public void run() {
21             System.out.println("thread2");
22         }
23     });
24 
25     static Thread thread3 = new Thread(new Runnable() {
26         @Override
27         public void run() {
28             System.out.println("thread3");
29         }
30     });
31 }

结果：

1 thread1
2 thread2
3 thread3

join方法的原理就是调用相应线程的wait方法进行等待操作的，例如A线程中调用了B线程的join方法，则相当于在A线程中调用了B线程的wait方法，当B线程执行完（或者到达等待时间），B线程会自动调用自身的notifyAll方法唤醒A线程，从而达到同步的目的。

源码：

 1 public final synchronized void join(long millis)
 2     throws InterruptedException {
 3         long base = System.currentTimeMillis();
 4         long now = 0;
 5 
 6         if (millis < 0) {
 7             throw new IllegalArgumentException("timeout value is negative");
 8         }
 9 
10         if (millis == 0) {
11             while (isAlive()) {
12                 wait(0);
13             }
14         } else {
15             while (isAlive()) {
16                 long delay = millis - now;
17                 if (delay <= 0) {
18                     break;
19                 }
20                 wait(delay);
21                 now = System.currentTimeMillis() - base;
22             }
23         }
24     }
wait(0)是当前主线程进入等待

 

方案二：
 1 public class Test {
 2     public static void main(String[] args) throws InterruptedException {
 3         executorService.submit(thread1);
 4         executorService.submit(thread2);
 5         executorService.submit(thread3);
 6     }
 7 
 8     static ExecutorService executorService = Executors.newSingleThreadExecutor();
 9 
10     static Thread thread1 = new Thread(new Runnable() {
11         @Override
12         public void run() {
13             System.out.println("thread1");
14         }
15     });
16 
17     static Thread thread2 = new Thread(new Runnable() {
18         @Override
19         public void run() {
20             System.out.println("thread2");
21         }
22     });
23 
24     static Thread thread3 = new Thread(new Runnable() {
25         @Override
26         public void run() {
27             System.out.println("thread3");
28         }
29     });
30 }

执行结果：

thread1
thread2
thread3

单线程化线程池(newSingleThreadExecutor)的优点，串行执行所有任务

 
posted @ 2018-11-19 09:14  jessee-blog  阅读(256)  评论(0)  编辑  收藏  举报