Multi-Programming-15 线程顺序执行
来源:互联网 发布:淘宝和易趣的相同点 编辑:程序博客网 时间:2024/06/05 14:26
1. 问题
Q:假设有三个线程,假设为t1, t2, t3, 如何保证线程的顺序执行t1->t2->t3?
A:首先,java语言多线程机制并没有执行顺序的属性;其次,设置线程的优先级并不能确保线程的准确执行顺序。所以,这里需要采用其他方式保证执行了。
本文提供了三种解决方式:
1). ReentrantLock: 重入锁方案。
2). Condition: 类似于wait(), notify()
3). Semaphore: 利用互斥信号量解决。
2. 解决思路
2.1 ReentrantLock机制
package com.fqyuan._10sequenceExecute;import java.util.concurrent.locks.ReentrantLock;public class SequenceStop { private static ReentrantLock lock = new ReentrantLock(); private /* volatile */ static int state = 0; public static void main(String[] args) { new Thread1().start(); new Thread2().start(); new Thread3().start(); } static class Thread1 extends Thread { /* * 主要有以下问题: 1)使用了int变量计数保证了线程的执行顺序; * 2)每次for循环之后的变量更新条件是在if语句中变化的,如果state条件没有满足则线程释放所获取的锁对象。 * 3)因为对于state的访问是在锁获取之间进行的,可以保证是同步访问的,故而可以不使用volatile关键字。 */ @Override public void run() { for (int i = 0; i < 10;) { lock.lock(); if (state % 3 == 0) { try { Thread.sleep(500); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } System.out.print("A"); state++; i++; } lock.unlock(); } } } static class Thread2 extends Thread { @Override public void run() { for (int i = 0; i < 10;) { lock.lock(); if (state % 3 == 1) { try { Thread.sleep(500); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } System.out.print("B"); state++; i++; } lock.unlock(); } } } static class Thread3 extends Thread { @Override public void run() { for (int i = 0; i < 10;) { lock.lock(); if (state % 3 == 2) { try { Thread.sleep(500); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } System.out.print("C"); state++; i++; } lock.unlock(); } } }}
2.2 Condiotn
package com.fqyuan._10sequenceExecute;import java.util.concurrent.locks.Condition;import java.util.concurrent.locks.ReentrantLock;public class SequenceStop1 { private static ReentrantLock lock = new ReentrantLock(); private static Condition A = lock.newCondition(); private static Condition B = lock.newCondition(); private static Condition C = lock.newCondition(); private static int count = 0; public static void main(String[] args) { new Thread1().start(); new Thread2().start(); // 因为主线程和用户线程是并发执行的,如果没有thread.join()语句,则打印的结果将为0 Thread thread3 = new Thread3(); thread3.start(); try { thread3.join(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } System.out.println(count); } static class Thread1 extends Thread { /* * 和纯粹使用ReentrantLock类似,都是用了一个int类型量决定线程的执行顺序。 * 不过使用了Conditon之后更加灵活。注意以下问题: 1) 使用了while语句判别wait条件是为了防止spurious * wakeup; 2) 在调用了await之后,线程将放弃当前的lock; 3)signal将会唤醒等待在该condition的线程。 */ @Override public void run() { try { lock.lock(); for (int i = 0; i < 10; i++) { while (count % 3 != 0) A.await(); // Release the lock it owns now! Thread.sleep(500); System.out.print("A"); count++; B.signal(); } } catch (InterruptedException e) { e.printStackTrace(); } finally { lock.unlock(); } } } static class Thread2 extends Thread { @Override public void run() { lock.lock(); try { for (int i = 0; i < 10; i++) { while (count % 3 != 1) B.await(); Thread.sleep(500); System.out.print("B"); count++; C.signal(); } } catch (InterruptedException e) { e.printStackTrace(); } finally { lock.unlock(); } } } static class Thread3 extends Thread { @Override public void run() { lock.lock(); try { for (int i = 0; i < 10; i++) { while (count % 3 != 2) C.await(); Thread.sleep(500); System.out.print("C"); count++; A.signal(); } } catch (InterruptedException e) { e.printStackTrace(); } finally { lock.unlock(); } } }}
2.3 使用互斥信号量
相对于使用ReentrantLock而言整体上较为简单,无须设置保证顺序的整形变量。但是在运行时需注意将非首先执行的线程信号量的availablePermits 置为0.
信号量的acquire() & release()是环形进行的,这样保证了执行的顺序!
package com.fqyuan._10sequenceExecute;import java.util.concurrent.Semaphore;public class SequenceStop2 { private static Semaphore A = new Semaphore(1); private static Semaphore B = new Semaphore(1); private static Semaphore C = new Semaphore(1); public static void main(String[] args) throws InterruptedException { B.acquire(); C.acquire(); new ThreadA().start(); new ThreadB().start(); new ThreadC().start(); } static class ThreadA extends Thread { @Override public void run() { try { for (int i = 0; i < 10; i++) { A.acquire(); System.out.print("A"); Thread.sleep(500); B.release(); } } catch (InterruptedException e) { } } } static class ThreadB extends Thread { @Override public void run() { try { for (int i = 0; i < 10; i++) { B.acquire(); System.out.print("B"); Thread.sleep(500); C.release(); } } catch (InterruptedException e) { } } } static class ThreadC extends Thread { @Override public void run() { try { for (int i = 0; i < 10; i++) { C.acquire(); System.out.print("C"); Thread.sleep(500); A.release(); } } catch (InterruptedException e) { } } }}
//Running result:ABCABCABCABCABCABCABCABCABCABC
3. 说明
本文内容参考自此处,可自行查看。
阅读全文
0 0
- Multi-Programming-15 线程顺序执行
- Multi-Programming-14 线程中断
- Multi-Programming-16 线程死锁实现方式
- 线程顺序执行(phtread)
- 控制线程顺序执行
- 线程执行顺序例子
- 三个线程顺序执行
- thread 线程执行顺序
- Multi-Programming-8 线程安全类实现生产者和消费者
- Java 线程的执行顺序
- JAVA线程分组顺序执行
- CountDownLatch控制线程执行顺序
- java 线程按顺序执行
- Android中让多个线程顺序执行
- NSOperation 线程终止 顺序执行
- Java线程循环顺序执行
- 三个线程依次顺序执行
- 控制线程的执行顺序
- 【17/8】ui复习之togglebutton
- 无线电项目 js总结
- POJ3070-Fibonacci
- 背包问题小结
- 卷积神经网络
- Multi-Programming-15 线程顺序执行
- 输出hello world
- javascript分享给朋友插件
- 【算法】n根彩色粉笔,m根白色粉笔,粉笔组合,求利益最大化
- 【POJ 1995】Raising Modulo Numbers(快速幂)
- 学习笔记---css中清除浮动
- Laravel CSRF
- 大整数类
- 历年阿里面试题汇总(2017年不断更新中)