❌ TODO：本文仅供参考，极有可能有误

1.生产者消费者问题（信号量）

参考教材中的生产者消费者算法，创建5个进程，其中两个进程为生产者进程，3个进程为消费者进程。一个生产者进程试图不断地在一个缓冲中写入大写字母，另一个生产者进程试图不断地在缓冲中写入小写字母。3个消费者不断地从缓冲中读取一个字符并输出。为了使得程序的输出易于看到结果，仿照的实例程序，分别在生产者和消费者进程的合适的位置加入一些随机睡眠时间。

💖 ProducerConsumerExample.java

import java.util.Random;
import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;public class ProducerConsumerExample {// 定义缓冲区的大小为10private static final int BUFFER_SIZE = 10;// 创建一个阻塞队列，用于存放字符，大小为BUFFER_SIZEprivate static final BlockingQueue<Character> buffer = new ArrayBlockingQueue<>(BUFFER_SIZE);// 定义生产者类，继承自Threadstatic class Producer extends Thread {// 是否生成大写字母private final boolean uppercase;// 随机数生成器private final Random random = new Random();// 构造函数，接收一个布尔值，决定生成大写还是小写字母public Producer(boolean uppercase) {this.uppercase = uppercase;}// 重写run方法，定义生产者的行为@Overridepublic void run() {try {// 循环直到线程被中断while (!Thread.currentThread().isInterrupted()) {// 根据是否生成大写字母，生成随机字符char item = uppercase ? Character.toUpperCase((char) ('A' + random.nextInt(26))) :Character.toLowerCase((char) ('a' + random.nextInt(26)));// 将生成的字符放入缓冲区buffer.put(item);// 打印生产信息System.out.println("Produced " + item + " by " + this.getName());// 随机休眠一段时间Thread.sleep(random.nextInt(1000));}} catch (InterruptedException e) {// 捕获InterruptedException异常，并重新中断当前线程Thread.currentThread().interrupt();}}}// 定义消费者类，继承自Threadstatic class Consumer extends Thread {// 随机数生成器private final Random random = new Random();// 重写run方法，定义消费者的行为@Overridepublic void run() {try {// 循环直到线程被中断while (!Thread.currentThread().isInterrupted()) {// 从缓冲区取出一个字符char item = (char) buffer.take();// 打印消费信息System.out.println("Consumed " + item + " by " + this.getName());// 随机休眠一段时间Thread.sleep(random.nextInt(1000));}} catch (InterruptedException e) {// 捕获InterruptedException异常，并重新中断当前线程Thread.currentThread().interrupt();}}}// 主函数，程序的入口点public static void main(String[] args) {// 创建两个生产者线程，一个生成大写字母，一个生成小写字母Producer producer1 = new Producer(true);Producer producer2 = new Producer(false);// 创建三个消费者线程Consumer consumer1 = new Consumer();Consumer consumer2 = new Consumer();Consumer consumer3 = new Consumer();// 启动所有生产者和消费者线程producer1.start();producer2.start();consumer1.start();consumer2.start();consumer3.start();try {// 等待所有线程完成producer1.join();producer2.join();consumer1.join();consumer2.join();consumer3.join();} catch (InterruptedException e) {// 打印异常信息e.printStackTrace();}}
}

可选的实验：在上面实验的基础上实现部分消费者有选择地消费某些产品。例如一个消费者只消费小写字符，一个消费者只消费大写字母，而另一个消费者则无选择地消费任何产品。消费者要消费的产品没有时，消费者进程被阻塞。注意缓冲的管理。

🏆 运行结果

💖 ProducerConsumerSelectiveExample.java

import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
import java.util.Random;
import java.util.concurrent.TimeUnit;public class ProducerConsumerSelectiveExample {private static final int BUFFER_SIZE = 10;private static final BlockingQueue<Character> buffer = new ArrayBlockingQueue<>(BUFFER_SIZE);static class Producer extends Thread {private final boolean uppercase;private final Random random = new Random();public Producer(boolean uppercase) {this.uppercase = uppercase;}@Overridepublic void run() {try {while (!Thread.currentThread().isInterrupted()) {char item = uppercase ? Character.toUpperCase((char) ('A' + random.nextInt(26))) :Character.toLowerCase((char) ('a' + random.nextInt(26)));buffer.put(item);System.out.println("Produced " + item + " by " + this.getName());Thread.sleep(random.nextInt(1000));}} catch (InterruptedException e) {Thread.currentThread().interrupt();}}}static class Consumer extends Thread {private final boolean onlyUppercase;private final Random random = new Random();public Consumer(boolean onlyUppercase) {this.onlyUppercase = onlyUppercase;}@Overridepublic void run() {try {while (!Thread.currentThread().isInterrupted()) {Character item =  buffer.poll(100, TimeUnit.MILLISECONDS); // 等待最多100毫秒if (item ==null) {// 如果没有找到想要的产品，跳过此次循环continue;}// 根据消费者类型进行选择性消费boolean consume = onlyUppercase ? Character.isUpperCase(item) :!onlyUppercase || Character.isLowerCase(item);if (consume) {buffer.remove(item); // 从队列中移除消费的项System.out.println("Consumed " + item + " by " + this.getName());}// 这里不需要 else 块，因为如果 consume 为 false，item 已经被检查过不是所需类型// 并且已经被忽略，不需要再次检查或睡眠Thread.sleep(random.nextInt(1000));}} catch (InterruptedException e) {Thread.currentThread().interrupt();}}}public static void main(String[] args) {Producer producer1 = new Producer(true);Producer producer2 = new Producer(false);Consumer consumerUppercaseOnly = new Consumer(true); // 只消费大写字母Consumer consumerLowercaseOnly = new Consumer(false); // 只消费小写字母Consumer consumerAny = new Consumer(true); // 消费任何产品producer1.start();producer2.start();consumerUppercaseOnly.start();consumerLowercaseOnly.start();consumerAny.start();try {producer1.join();producer2.join();consumerUppercaseOnly.join();consumerLowercaseOnly.join();consumerAny.join();} catch (InterruptedException e) {e.printStackTrace();}}
}

🏆 运行结果

2.实现睡觉的理发师问题

（同步互斥方式采用信号量或mutex方式均可）
理发师问题的描述：一个理发店接待室有n张椅子，工作室有1张椅子；没有顾客时，理发师睡觉；第一个顾客来到时，必须将理发师唤醒；顾客来时如果还有空座的话，他就坐在一个座位上等待；如果顾客来时没有空座位了，他就离开，不理发了；当理发师处理完所有顾客，而又没有新顾客来时，他又开始睡觉。

💖 BarberShop.java

import java.util.concurrent.Semaphore;
import java.util.Random;public class BarberShop {private final int capacity;private final Semaphore barber = new Semaphore(1);private final Semaphore chair = new Semaphore(0);private final Semaphore[] seats;private final Random random = new Random();public BarberShop(int capacity) {this.capacity = capacity;this.seats = new Semaphore[capacity];for (int i = 0; i < capacity; i++) {this.seats[i] = new Semaphore(1);}}public void barberAction() {try {while (!Thread.currentThread().isInterrupted()) {barber.acquire();chair.acquire(); // 理发师等待顾客坐下System.out.println("理发师醒来开始理发。");// 模拟随机理发时间Thread.sleep(random.nextInt(1000) + 100);System.out.println("理发师完成理发。");barber.release(); // 理发师完成服务，可以服务下一个顾客}} catch (InterruptedException e) {Thread.currentThread().interrupt();}}public void customerAction() {try {while (!Thread.currentThread().isInterrupted()) {boolean foundSeat = false;for (Semaphore seat : seats) {if (seat.tryAcquire()) {foundSeat = true;System.out.println("顾客坐在椅子上等待理发。");chair.release(); // 唤醒理发师// 模拟理发师服务时间Thread.sleep(random.nextInt(1000) + 100);System.out.println("顾客离开理发椅。");seat.release(); // 顾客离开，释放椅子break;}}if (!foundSeat) {System.out.println("没有空椅子，顾客离开。");}// 模拟随机顾客到来时间Thread.sleep(random.nextInt(1000) + 500);}} catch (InterruptedException e) {Thread.currentThread().interrupt();}}public static void main(String[] args) {final int CAPACITY = 3; // 假设有3张椅子BarberShop shop = new BarberShop(CAPACITY);Thread barberThread = new Thread(() -> shop.barberAction());barberThread.start();// 创建顾客线程Thread[] customerThreads = new Thread[5];for (int i = 0; i < customerThreads.length; i++) {customerThreads[i] = new Thread(() -> shop.customerAction());customerThreads[i].start();}}
}

🏆 运行结果