实现多线程的三种方式:继承Thread类、实现Runnable接口、使用Callable和Future接口、使用线程池创建线程
一、继承Thread类,重写run方法
public class MyThread extends Thread{ @Override public void run() { for (int i = 0; i < 20; i++) { System.out.println("我在听音乐"); } } public static void main(String[] args) { MyThread myThread = new MyThread(); myThread.start(); for (int i = 0; i < 20; i++) { System.out.println("我在学习多线程"); } } }
二、实现Runnable接口
public class MyThread2 implements Runnable{ @Override public void run() { for (int i = 0; i < 20; i++) { System.out.println("我在听音乐"); } } public static void main(String[] args) { Thread myThread = new Thread(new MyThread2()); myThread.start(); for (int i = 0; i < 20; i++) { System.out.println("我在学习多线程"); } } }
三、使用Callable和Future接口创建线程
public class MyThread4 { public static void main(String[] args) throws ExecutionException, InterruptedException { Test test = new Test(); FutureTask<Integer> ft = new FutureTask<>(test); Thread thread = new Thread(ft); thread.start(); for (int i = 0; i < 200; i++) { System.out.println("我在学习多线程"); } //获取返回值 ft.get(); } static class Test implements Callable<Integer>{ @Override public Integer call() throws Exception { for (int i = 0; i < 200; i++) { System.out.println("我在听音乐"); } return null; } } }
四、使用线程池创建线程
实现Callable接口,需要返回值类型
重写call方法,需要抛出异常
创建目标对象
创建执行服务:ExecutorService ser = Executors.newFixedThreadPool(3);
提交执行:Futuresub1 = ser.submit(test1);
获取结果:Integer integer1 = sub1.get();
关闭服务:ser.shutdownNow();
public class MyThread3 { public static void main(String[] args) throws ExecutionException, InterruptedException { Test test1 = new Test(1); Test test2 = new Test(2); Test test3 = new Test(3); //创建执行服务 ExecutorService ser = Executors.newFixedThreadPool(3); //提交执行 Future<Integer> sub1 = ser.submit(test1); Future<Integer> sub2 = ser.submit(test2); Future<Integer> sub3 = ser.submit(test3); //获取结果 Integer integer1 = sub1.get(); Integer integer2 = sub2.get(); Integer integer3 = sub3.get(); //关闭服务 ser.shutdownNow(); } static class Test implements Callable<Integer>{ private int a; public Test(int a) { this.a = a; } @Override public Integer call() throws Exception { for (int i = 0; i < 20; i++) { System.out.println("我在听音乐"+a); } return null; } } }