Java 多线程的两种简单实现方法

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简述：

1. 简单用一下Java的多线程的两种粗略的实现方法

继承Thread和实现Runnable接口

实现：

1. 随机给一个自然数n，分发给多个线程，每个线程计算n项的Fibonacci数列的和，计算完成后输出，

1）继承Thread类实现，并重写run()方法

package test.multithread.Fibonacci;import java.util.Random;public class TestA_1 extends Thread{private int n;private static int taskCount = 0;private final int id = taskCount++;public TestA_1(int n) {this.n = n;}public static void main(String[] args){for(int i = 0;i < 10;i++){Random random = new Random();  Integer x = random.nextInt(100);new TestA_1(x).start();}}@Override//run implement the counting of different n public void run() {int sum = 0;if(n == 0 || n == 1){sum += n;}else{sum = 1;for(int i = 1;i < n;i++){sum += i;}}System.out.println("id: " + id + ",\t" + n + ": " + sum);Thread.yield(); // now the CPU could transfer the thread to a new Thread}}

2）实现Runnable接口中run方法

package test.multithread.Fibonacci;import java.util.Random;public class TestA_2 implements Runnable{private int n;private static int taskCount = 0;private final int id = taskCount++;public TestA_2(int n) {this.n = n;}public static void main(String[] args){for(int i = 0;i < 10;i++){Random random = new Random();  Integer x = random.nextInt(100);new TestA_2(x).run();}}@Override//run implement the counting of different n public void run() {int sum = 0;if(n == 0 || n == 1){sum += n;}else{sum = 1;for(int i = 1;i < n;i++){sum += i;}}System.out.println("id: " + id + ",\t" + n + ": " + sum);Thread.yield(); // now the CPU could transfer the thread to a new Thread}}

结论，从这个测试上看两者在多线程的实现上都可以完成，任务的分发，目前没什么区别

两者输出相似都如：id是线程的id，后面的数字是Fibonacci数列n个项的和