Java 多线程的两种简单实现方法

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