用P、V操作实现哲学家进餐问题的第2种解法
时间: 2024-05-24 18:13:26 浏览: 46
可以使用信号量(Semaphore)实现哲学家进餐问题的第2种解法。具体实现方式是,每个哲学家都有一个信号量,表示他的左右两边的餐叉是否可用。当哲学家想要进餐时,他必须先获取他左右两边的餐叉,如果餐叉都可用,则他获取到信号量并开始进餐,否则他会等待直到餐叉可用。当他吃完后,他会释放他左右两边的餐叉的信号量,使得其他哲学家可以使用这些餐叉。
相关问题
用java实现实现哲学家进餐问题
哲学家进餐问题是一个经典的同步问题,主要涉及到多个进程(哲学家)之间的资源竞争问题。在这个问题中,每个哲学家都需要交替地进行思考和进餐,但是他们共享一些资源,如餐具等,因此需要协调它们的行为以避免死锁和饥饿等问题。
下面是使用 Java 实现哲学家进餐问题的示例代码:
```java
import java.util.concurrent.Semaphore;
public class Philosopher implements Runnable {
private final int id;
private final Semaphore leftFork;
private final Semaphore rightFork;
public Philosopher(int id, Semaphore leftFork, Semaphore rightFork) {
this.id = id;
this.leftFork = leftFork;
this.rightFork = rightFork;
}
@Override
public void run() {
try {
while (true) {
think();
leftFork.acquire();
rightFork.acquire();
eat();
rightFork.release();
leftFork.release();
}
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
return;
}
}
private void think() throws InterruptedException {
System.out.println("Philosopher " + id + " is thinking");
Thread.sleep((long) (Math.random() * 5000));
}
private void eat() throws InterruptedException {
System.out.println("Philosopher " + id + " is eating");
Thread.sleep((long) (Math.random() * 5000));
}
}
public class DiningPhilosophers {
private static final int NUM_PHILOSOPHERS = 5;
public static void main(String[] args) {
Semaphore[] forks = new Semaphore[NUM_PHILOSOPHERS];
for (int i = 0; i < NUM_PHILOSOPHERS; i++) {
forks[i] = new Semaphore(1);
}
Philosopher[] philosophers = new Philosopher[NUM_PHILOSOPHERS];
for (int i = 0; i < NUM_PHILOSOPHERS; i++) {
philosophers[i] = new Philosopher(i, forks[i], forks[(i + 1) % NUM_PHILOSOPHERS]);
new Thread(philosophers[i]).start();
}
}
}
```
在这个示例代码中,我们创建了一个 `Philosopher` 类和一个 `DiningPhilosophers` 类,其中 `Philosopher` 类实现了哲学家的行为,而 `DiningPhilosophers` 类则负责创建和启动哲学家线程。
在 `Philosopher` 类中,我们使用了两个信号量 `leftFork` 和 `rightFork` 表示哲学家左边和右边的餐叉。在 `run` 方法中,哲学家先思考一段时间,然后尝试获取左右餐叉的信号量。如果成功获取了两个餐叉的信号量,哲学家就可以进餐。最后,哲学家释放左右餐叉的信号量,以便其他哲学家可以使用它们。
在 `DiningPhilosophers` 类中,我们首先创建了一个 `Semaphore` 数组 `forks`,其中每个元素都表示一个餐叉。然后,我们创建了五个 `Philosopher` 对象,并将它们的左右餐叉设置为相邻的两个 `Semaphore` 对象。最后,我们创建了五个线程,每个线程都对应一个哲学家,并启动它们。当程序运行时,每个哲学家会交替进行思考和进餐,直到程序被中断或终止。
操作系统哲学家进餐问题java源码实现
好的,这里是操作系统哲学家进餐问题的Java程序实现:
```java
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class DiningPhilosophers {
private static final int NUMBER_OF_PHILOSOPHERS = 5;
private static final int SIMULATION_TIME = 1000;
public static void main(String[] args) throws InterruptedException {
Philosopher[] philosophers = new Philosopher[NUMBER_OF_PHILOSOPHERS];
Object[] forks = new Object[NUMBER_OF_PHILOSOPHERS];
Lock[] locks = new Lock[NUMBER_OF_PHILOSOPHERS];
for (int i = 0; i < NUMBER_OF_PHILOSOPHERS; i++) {
forks[i] = new Object();
locks[i] = new ReentrantLock();
}
for (int i = 0; i < NUMBER_OF_PHILOSOPHERS; i++) {
philosophers[i] = new Philosopher(i, forks[i], forks[(i + 1) % NUMBER_OF_PHILOSOPHERS], locks[i],
locks[(i + 1) % NUMBER_OF_PHILOSOPHERS]);
new Thread(philosophers[i]).start();
}
Thread.sleep(SIMULATION_TIME);
for (Philosopher philosopher : philosophers) {
philosopher.setFull(true);
}
}
}
class Philosopher implements Runnable {
private final int id;
private final Object leftFork;
private final Object rightFork;
private final Lock leftLock;
private final Lock rightLock;
private volatile boolean isFull = false;
public Philosopher(int id, Object leftFork, Object rightFork, Lock leftLock, Lock rightLock) {
this.id = id;
this.leftFork = leftFork;
this.rightFork = rightFork;
this.leftLock = leftLock;
this.rightLock = rightLock;
}
@Override
public void run() {
try {
while (!isFull) {
think();
pickUpForks();
eat();
putDownForks();
}
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
return;
}
}
private void think() throws InterruptedException {
System.out.println("Philosopher " + id + " is thinking");
Thread.sleep((long) (Math.random() * 10000));
}
private void pickUpForks() throws InterruptedException {
leftLock.lock();
try {
System.out.println("Philosopher " + id + " picked up left fork");
rightLock.lock();
try {
System.out.println("Philosopher " + id + " picked up right fork");
} catch (Exception e) {
leftLock.unlock();
throw e;
}
} catch (Exception e) {
rightLock.unlock();
throw e;
}
}
private void eat() throws InterruptedException {
System.out.println("Philosopher " + id + " is eating");
Thread.sleep((long) (Math.random() * 10000));
}
private void putDownForks() {
leftLock.unlock();
rightLock.unlock();
System.out.println("Philosopher " + id + " put down forks");
}
public void setFull(boolean isFull) {
this.isFull = isFull;
}
}
```
这个程序中,有5个哲学家和5个餐叉,每个哲学家需要同时拿到左右两边的餐叉才能进餐,否则需要等待。使用 `Lock` 和 `ReentrantLock` 来实现锁定餐叉。程序的运行时间为1000毫秒,每个哲学家会先思考一段时间,然后尝试拿起餐叉,进餐,然后放下餐叉,最后再开始思考。
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