Java 多線程的兩種簡單實現方法__Java

簡述：

1. 簡單用一下Java的多線程的兩種粗略的實現方法

繼承Thread和實現Runnable介面

實現：

1. 隨機給一個自然數n， 分發給多個線程， 每個線程計算n項的Fibonacci數列的和，計算完成後輸出，

1）繼承Thread類實現，並重寫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）實現Runnable介面中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}}

結論，從這個測試上看兩者在多線程的實現上都可以完成，任務的分發，目前沒什麼區別

兩者輸出相似都如：id是線程的id，  後面的數字是Fibonacci數列n個項的和