java線程推塔代碼 java線程死鎖代碼

java線程的經典代碼

package threadgroup;

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class ThreadDemo3 extends Thread {

private String name;

private int delay;

public ThreadDemo3(String sname, int i_delay) {

name = sname;

delay = i_delay;

}

public void run() {

try {

sleep(delay);

} catch (InterruptedException e) {

}

System.out.println("多線程測試！\n" + name + "\n" + delay);

}

}

public class testMyThread {

public static void main(String[] args) {

ThreadDemo3 th1,th2,th3;

th1 = new ThreadDemo3("線程1", (int) (Math.random() * 900));

th2 = new ThreadDemo3("線程2", (int) (Math.random() * 900));

th3 = new ThreadDemo3("線程3", (int) (Math.random() * 900));

th1.start();

th2.start();

th3.start();

}

}

package threadgroup;

public class threadDemo {

public static void main(String[] args) {

Thread t = Thread.currentThread();

t.setName("你好嗎？");

System.out.println("正在進行的Thread是：" + t);

try {

for (int i = 0; i 5; i++) {

System.out.println("我不叫穆繼超" + i);

Thread.sleep(3000);

}

} catch (Exception e) {

// TODO: handle exception

System.out.println("Thread has wrong" + e.getMessage());

}

}

}

package threadgroup;

public class threadDemo2 implements Runnable {

public threadDemo2() {

Thread t1 = Thread.currentThread();

t1.setName("第一個主進程");

System.out.println("正在運行" + t1);

Thread t2 = new Thread(this, "");

System.out.println("在創建一個進程");

t2.start();

try {

System.out.println("使他進入第一個睡眠狀態");

Thread.sleep(2000);

} catch (InterruptedException e) {

System.out.println("Thread has wrong" + e.getMessage());

}

System.out.println("退出第一個進程");

}

public void run() {

try {

for (int i = 0; i 5; i++) {

System.out.println("進程" + i);

Thread.sleep(3000);

}

} catch (InterruptedException e) {

// TODO: handle exception

System.out.println("Thread has wrong" + e.getMessage());

}

System.out.println("退出第二個進程");

}

public static void main(String[] args) {

new threadDemo2();

}

}

求一個JAVA多線程例子，最好有代碼，謝謝啦！

package a.b.test;

import java.util.Date;

import java.util.concurrent.Callable;

import java.util.concurrent.ExecutorService;

import java.util.concurrent.Executors;

import java.util.concurrent.Future;

public class Calculate1000 implements CallableInteger{

public Calculate1000(){}

public Calculate1000(int a, int b){

this.a = a;

this.b = b;

}

int a;

int b;

/**

* @param args

* @throws Exception

*/

public static void main(String[] args) throws Exception {

//同步

Calculate1000 ca1 = new Calculate1000();

Date ds1 = new Date();

int result = 0;

for(int i = 1 ; i = 1000 ; i++){

result = ca1.add(i, result);

}

System.out.println(result);

System.out.println("同步用時" + (new Date().getTime() - ds1.getTime()) + "MS");

//異步

Date ds2 = new Date();

result = 0;

ExecutorService es = Executors.newFixedThreadPool(2);

FutureInteger future1 = es.submit(new Calculate1000(1,500));

FutureInteger future2 = es.submit(new Calculate1000(501,1000));

result = future1.get() + future2.get();

System.out.println(result);

System.out.println("異步用時" + (new Date().getTime() - ds2.getTime()) + "MS");

es.shutdown();

}

private int add(int a, int b) throws Exception{

Thread.sleep(10);

return a + b;

}

@Override

public Integer call() throws Exception {

int res = 0;

for(int i = a ; i = b ; i++){

res = this.add(res, i);

}

return res;

}

}

樓主你試一下這段代碼行不行，行的話請采納！

java多線程編程代碼如下，輸出結果如下:

首先，你同步的是具體的某個Test實例， 對于那個實例來說，實際上只有一個線程訪問了那個代碼塊，但是sum和other卻是多個線程同時去進行訪問，實際上這是不安全的，如果你想實現每次都輸出10000的效果，那么正確的應該是在Test.class上加鎖，而不是獲取Test實例的鎖，修改后的代碼如下：

public?class?Test?extends?Thread?{

public?static?int?sum?=?10000;

public?static?int?other?=?0;

public?void?getMoney()?{

synchronized?(Test.class)?{

System.out.println(Thread.currentThread().getName()?+?"?開始執行");

sum?=?sum?-?100;

System.out.println("sum-100");

other?=?other?+?100;

System.out.println("other+100");

System.out.println(sum?+?other);

System.out.println(Thread.currentThread().getName()?+?"?執行完成");

}

}

public?void?run()?{

getMoney();

}

public?static?void?main(String[]?agrs)?{

Thread?t[]?=?new?Thread[10];

for?(int?i?=?0;?i?=?9;?i++)?{

t[i]?=?new?Test();

t[i].start();

}

}

}

// 上面代碼能得到你的結果

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