java并发-如何保证线程按照顺序执行？

【readme】

使用只有单个线程的线程池（最简单）
Thread.join()
可重入锁 ReentrantLock + Condition 条件变量（多个）；原理如下：
1. 任务1执行前在锁1上阻塞；执行完成后在锁2上唤醒；
2. 任务2执行前在锁2上阻塞，执行完成后在锁3上唤醒；
3. 任务n执行前在锁n上阻塞，执行完成后在锁n+1上唤醒；
4. 以此类推 ..............
5. 补充：
  1. 第1条任务执行前可以不阻塞，但执行完成后必须唤醒；（如果要阻塞，则可以让主线程来唤醒第1条任务）；
  2. 补充：最后一条任务执行后可以不唤醒，但执行前必须阻塞；（如果要唤醒，则最后一条任务执行完成后唤醒主线程）；
与可重入锁类似，可以使用monitor监视器锁（多个）；
与可重入锁类似，使用 Semaphore 信号量（多个）；
与可重入锁类似，CountDownLatch ：倒计时锁存器（多个）；
与可重入锁类似，CyclicBarrier 循环栅栏（多个）；

【1】单个线程的线程池

参数设置：核心线程数=1，最大线程数=1，就能保证线程池中只有1个线程在运行；

public class OrderlySingleThreadPoolTest {public static void main(String[] args) {ThreadPoolExecutor singleThreadPool =new ThreadPoolExecutor(1, 1, 0, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<>(100));singleThreadPool.execute(new Task(1));singleThreadPool.execute(new Task(2));singleThreadPool.execute(new Task(3));singleThreadPool.execute(new Task(4));singleThreadPool.execute(new Task(5));singleThreadPool.shutdown();}private static class Task implements Runnable {int order; // 执行序号Task(int order) {this.order = order;}@Overridepublic void run() {try {TimeUnit.SECONDS.sleep(3);} catch (InterruptedException e) {throw new RuntimeException(e);}PrintUtils.print("序号=" + order + "执行完成");}}
}

【打印结果】

2024-06-09 16:07:59.398 序号=1执行完成
2024-06-09 16:08:02.404 序号=2执行完成
2024-06-09 16:08:05.411 序号=3执行完成
2024-06-09 16:08:08.425 序号=4执行完成
2024-06-09 16:08:11.439 序号=5执行完成

【2】thread.join()

main 调用 t1.join()，则main线程阻塞直到t1线程执行完成；如下。

public class ThreadJoinTest {public static void main(String[] args) {f1();PrintUtils.print("主线程结束");}public static void f1() {Thread t1 = new Thread(()->{try {TimeUnit.SECONDS.sleep(5);PrintUtils.print("t1线程结束");} catch (InterruptedException e) {throw new RuntimeException(e);}});t1.start();try {t1.join();} catch (InterruptedException e) {throw new RuntimeException(e);}}
}

2024-06-09 07:44:38.474 t1线程结束
2024-06-09 07:44:38.573 主线程结束

【3】可重入锁+条件变量实现多个线程顺序执行

1. 补充：

condition.await() 调用前需要获取锁；调用后释放锁（其他线程可以获取该锁，因此得名为可重入），但当前线程阻塞；

condition.signal() 调用前需要获取锁；调用后释放锁；

public class OrderlyReentrantLockTest {private static Condition[][] build(ReentrantLock reentrantLock, int num) {Condition[][] arr = new Condition[num][2];arr[0] = new Condition[]{null, reentrantLock.newCondition()};int i = 1;for (; i < num - 1; i++) {arr[i] = new Condition[]{arr[i - 1][1], reentrantLock.newCondition()};}arr[i] = new Condition[]{arr[i - 1][1], null};return arr;}public static void main(String[] args) {int threadNum = 5;ThreadPoolExecutor threadPool =new ThreadPoolExecutor(threadNum, threadNum, 0, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<>(100));ReentrantLock reentrantLock = new ReentrantLock(true);AtomicInteger unWaitNum = new AtomicInteger(threadNum);// 构建条件变量数组Condition[][] conditionTwoArr = build(reentrantLock, threadNum);// 提交任务for (int order = threadNum; order >= 1; order--) {OrderlyTask orderlyTask = new OrderlyTask(reentrantLock, conditionTwoArr[order - 1], order, unWaitNum);threadPool.execute(orderlyTask);// 阻塞成功，才提交下一个任务while (unWaitNum.get() == threadNum) ; // 这里可能死循环，但可以新增超时重试机制来处理PrintUtils.print("阻塞成功，线程order=" + order);} threadPool.shutdown();}private static class OrderlyTask implements Runnable {private ReentrantLock lock;private Condition[] conditions;private int order; // 执行序号private AtomicInteger unWaitNum;OrderlyTask(ReentrantLock reentrantLock, Condition[] conditions, int order, AtomicInteger unWaitNum) {this.lock = reentrantLock;this.conditions = conditions;this.order = order;this.unWaitNum = unWaitNum;}@Overridepublic void run() {lock.lock();try {unWaitNum.decrementAndGet();try {if (conditions[0] != null) {conditions[0].await(); // 在第1个条件变量上阻塞}} catch (Exception e) {unWaitNum.incrementAndGet();throw e;}// 处理业务逻辑TimeUnit.SECONDS.sleep(3);// 唤醒在第2个条件变量上阻塞的线程if (conditions[1] != null) {conditions[1].signal();}} catch (Exception e) {System.err.println(e);} finally {lock.unlock();}PrintUtils.print("执行完成， 线程order=" + order + ", 线程id=" + Thread.currentThread().getName());}}
}

打印结果：

2024-06-09 22:16:00.696 阻塞成功，线程order=5
2024-06-09 22:16:00.698 阻塞成功，线程order=4
2024-06-09 22:16:00.698 阻塞成功，线程order=3
2024-06-09 22:16:00.698 阻塞成功，线程order=2
2024-06-09 22:16:00.698 阻塞成功，线程order=1
2024-06-09 22:16:03.707 执行完成， 线程order=1, 线程id=pool-1-thread-5
2024-06-09 22:16:06.719 执行完成， 线程order=2, 线程id=pool-1-thread-4
2024-06-09 22:16:09.719 执行完成， 线程order=3, 线程id=pool-1-thread-3
2024-06-09 22:16:12.727 执行完成， 线程order=4, 线程id=pool-1-thread-2
2024-06-09 22:16:15.729 执行完成， 线程order=5, 线程id=pool-1-thread-1

【4】使用CountDownLatch倒计时锁存器

public class OrderlyCountDownLatchTest {private static CountDownLatch[][] build(int num) {CountDownLatch[][] arr = new CountDownLatch[num][2];arr[0] = new CountDownLatch[]{null, new CountDownLatch(1)};int i = 1;for (; i < num - 1; i++) {arr[i] = new CountDownLatch[]{arr[i - 1][1], new CountDownLatch(1)};}arr[i] = new CountDownLatch[]{arr[i - 1][1], null};return arr;}public static void main(String[] args) {int threadNum = 5;ThreadPoolExecutor threadPool =new ThreadPoolExecutor(threadNum, threadNum, 0, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<>(100));AtomicInteger unWaitNum = new AtomicInteger(threadNum);// 构建倒计时锁存器数组CountDownLatch[][] latchArr = build(threadNum);// 提交任务for (int order = threadNum; order >= 1; order--) {OrderlyTask orderlyTask = new OrderlyTask(latchArr[order - 1], order, unWaitNum);threadPool.execute(orderlyTask);// 阻塞成功，才提交下一个任务while (unWaitNum.get() == threadNum) ; // 这里可能死循环，但可以新增超时重试机制来处理PrintUtils.print("阻塞成功，线程order=" + order);}threadPool.shutdown();}private static class OrderlyTask implements Runnable {private CountDownLatch[] latchArr;private int order; // 执行序号private AtomicInteger unWaitNum;OrderlyTask(CountDownLatch[] latchArr, int order, AtomicInteger unWaitNum) {this.latchArr = latchArr;this.order = order;this.unWaitNum = unWaitNum;}@Overridepublic void run() {try {unWaitNum.decrementAndGet();try {if (latchArr[0] != null) {latchArr[0].await(); // 在第1个锁存器上阻塞}} catch (Exception e) {unWaitNum.incrementAndGet();throw e;}// 处理业务逻辑TimeUnit.SECONDS.sleep(3);// 唤醒在第2个条件变量上阻塞的线程if (latchArr[1] != null) {latchArr[1].countDown();}} catch (Exception e) {System.err.println(e);}PrintUtils.print("执行完成， 线程order=" + order + ", 线程id=" + Thread.currentThread().getName());}}
}

打印结果：

2024-06-09 22:35:13.648 阻塞成功，线程order=5
2024-06-09 22:35:13.651 阻塞成功，线程order=4
2024-06-09 22:35:13.651 阻塞成功，线程order=3
2024-06-09 22:35:13.651 阻塞成功，线程order=2
2024-06-09 22:35:13.651 阻塞成功，线程order=1
2024-06-09 22:35:16.664 执行完成， 线程order=1, 线程id=pool-1-thread-5
2024-06-09 22:35:19.676 执行完成， 线程order=2, 线程id=pool-1-thread-4
2024-06-09 22:35:22.682 执行完成， 线程order=3, 线程id=pool-1-thread-3
2024-06-09 22:35:25.684 执行完成， 线程order=4, 线程id=pool-1-thread-2
2024-06-09 22:35:28.688 执行完成， 线程order=5, 线程id=pool-1-thread-1