使用Thread.join(long million)方法，参数是毫秒
代码&解析如下：
解析：原本有t1和t2两个线程，根据实例化new Task（）时t1传入了4，t2传入了2，分别相当于t1需要执行4次，t2需要执行2次，但是在run方法中使用了Thread.sleep(1000)，所以t1执行4次就等于是执行4秒，t2同理。在t1.start启动后，调用了join方法设置了两秒的参数，相当于在t1执行两秒后就超时了，后面就是t1超时后的设置，
1、t1.interrupt()表示打断t1线程必须写，否则超时后还会继续执行。 2、run方法catch代码块中的return方法也是必须写的，否则t1线程也还会继续执行。
总结：Thread.join(long million)、nterrupt()、return三个地方需要同时书写，否则就会出现问题。
public class Test1 {public static void main(String[] args) {Task task1 = new Task("one", 4);Task task2 = new Task("two", 2);Thread t1 = new Thread(task1);Thread t2 = new Thread(task2);t1.start();try {t1.join(2000); // 在主线程中等待t1执行2秒} catch (InterruptedException e) {System.out.println("t1 interrupted when waiting join");e.printStackTrace();}t1.interrupt(); // 这里很重要，一定要打断t1,因为它已经执行了2秒。t2.start();try {t2.join(1000);} catch (InterruptedException e) {System.out.println("t2 interrupted when waiting join");e.printStackTrace();}}}class Task implements Runnable {public String name;private int time;public Task(String s, int t) {name = s;time = t;}public void run() {for (int i = 0; i < time; ++i) {System.out.println("task " + name + " " + (i + 1) + " round");try {Thread.sleep(1000);} catch (InterruptedException e) {System.out.println(name+ "is interrupted when calculating, will stop...");return; // 注意这里如果不return的话，线程还会继续执行，所以任务超时后在这里处理结果然后返回}}}}
t1超时后正确处理方式的打印结果：
如果调用过join()后没有书写interrupt()和return就会是以下错误打印：
因为设置了两秒后t1任务需要停止，所以正确的时候是不需要打印task one 3 round和task one 4 round的
Future.get(long million, TimeUnit unit) 配合Future.cancle(true)
f1.get(2, TimeUnit.SECONDS)就是设置两秒后超时，超时后会这个方法本身会抛出一个TimeoutException异常，有TimeoutException 这个异常时设置f1.cancel(true);，同样在call方法中也要进行return，不然超时后线程还依然会执行。
public class Test1 {static class Task implements Callable<Boolean> {public String name;private int time;public Task(String s, int t) {name = s;time = t;}@Overridepublic Boolean call() throws Exception {for (int i = 0; i < time; ++i) {System.out.println("task " + name + " round " + (i + 1));try {Thread.sleep(1000);} catch (InterruptedException e) {System.out.println(name+ " is interrupted when calculating, will stop...");return false; // 注意这里如果不return的话，线程还会继续执行，所以任务超时后在这里处理结果然后返回}}return true;}}public static void main(String[] args) {ExecutorService executor = Executors.newCachedThreadPool();Task task1 = new Task("one", 5);Future<Boolean> f1 = executor.submit(task1);try {if (f1.get(2, TimeUnit.SECONDS)) {// future将在2秒之后取结果System.out.println("one complete successfully");}} catch (InterruptedException e) {System.out.println("future在睡着时被打断");executor.shutdownNow();} catch (ExecutionException e) {System.out.println("future在尝试取得任务结果时出错");executor.shutdownNow();} catch (TimeoutException e) {System.out.println("future时间超时");f1.cancel(true);// executor.shutdownNow();// executor.shutdown();} finally {executor.shutdownNow();}}}
设置2秒后超时的运行结果：
ExecutorService.awaitTermination(long million, TimeUnit unit)
这个方法会一直等待所有的任务都结束，或者超时时间到立即返回，若所有任务都完成则返回true，否则返回false
代码如下：
public class Test1 {static class Task implements Runnable {public String name;private int time;public Task(String s, int t) {name = s;time = t;}public void run() {for (int i = 0; i < time; ++i) {try {Thread.sleep(1000);} catch (InterruptedException e) {System.out.println(name+ " is interrupted when calculating, will stop...");return; // 注意这里如果不return的话，线程还会继续执行，所以任务超时后在这里处理结果然后返回}System.out.println("task " + name + " " + (i + 1) + " round");}System.out.println("task " + name + " finished successfully");}}public static void main(String[] args) {ExecutorService executor = Executors.newCachedThreadPool();Task task = new Task("one", 5);Task task2 = new Task("two", 2);Future<?> future = executor.submit(task);Future<?> future2 = executor.submit(task2);List<Future<?>> futures = new ArrayList<Future<?>>();futures.add(future);futures.add(future2);try {if (executor.awaitTermination(3, TimeUnit.SECONDS)) {System.out.println("task finished");} else {System.out.println("task time out,will terminate");for (Future<?> f : futures) {if (!f.isDone()) {f.cancel(true);}}}} catch (InterruptedException e) {System.out.println("executor is interrupted");} finally {executor.shutdown();}}}
设置3秒超时，task2线程总共运行2秒，不存在超时情况，所以会返回task two finished successfully，task线程总共运行5秒，在第三秒超时，所以task超时后的结果都不会打印，而是提示task time out,will terminate。
设置一个守护线程，守护线程先sleep一段定时时间（睡眠时长就是超时时长），睡醒后打断它所监视的线程
public class Test1 {static class Task implements Runnable {private String name;private int time;public Task(String s, int t) {name = s;time = t;}public int getTime() {return time;}public void run() {for (int i = 0; i < time; ++i) {try {Thread.sleep(1000);} catch (InterruptedException e) {System.out.println(name+ " is interrupted when calculating, will stop...");return; // 注意这里如果不return的话，线程还会继续执行，所以任务超时后在这里处理结果然后返回}System.out.println("task " + name + " " + (i + 1) + " round");}System.out.println("task " + name + " finished successfully");}}static class Daemon implements Runnable {List<Runnable> tasks = new ArrayList<Runnable>();private Thread thread;private int time;public Daemon(Thread r, int t) {thread = r;time = t;}public void addTask(Runnable r) {tasks.add(r);}@Overridepublic void run() {while (true) {try {Thread.sleep(time * 1000);} catch (InterruptedException e) {e.printStackTrace();}thread.interrupt();}}}public static void main(String[] args) {Task task1 = new Task("one", 5);Thread t1 = new Thread(task1);Daemon daemon = new Daemon(t1, 4);Thread daemoThread = new Thread(daemon);daemoThread.setDaemon(true);t1.start();daemoThread.start();}}
设置4秒醒来，从而打断所监视的目标线程。运行结果：
