用代码解决上述问题
时间: 2024-02-18 11:02:01 浏览: 17
下面是一个使用Java语言实现的解决方案:
```java
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class Philosopher implements Runnable {
private final int id;
private final Lock leftChopstick;
private final Lock rightChopstick;
public Philosopher(int id, Lock leftChopstick, Lock rightChopstick) {
this.id = id;
this.leftChopstick = leftChopstick;
this.rightChopstick = rightChopstick;
}
@Override
public void run() {
while (true) {
try {
// 1. 拿起左边的筷子
leftChopstick.lock();
System.out.println("Philosopher " + id + " picks up left chopstick");
try {
// 2. 尝试拿起右边的筷子
if (rightChopstick.tryLock()) {
System.out.println("Philosopher " + id + " picks up right chopstick and starts eating");
// 3. 就餐一段时间
Thread.sleep(1000);
} else {
System.out.println("Philosopher " + id + " fails to pick up right chopstick and puts down left chopstick");
}
} finally {
// 4. 放下右边的筷子
rightChopstick.unlock();
}
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
return;
} finally {
// 5. 放下左边的筷子
leftChopstick.unlock();
}
// 6. 等待一段时间再重新尝试就餐
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
Thread.currentThread().interrupt();
return;
}
}
}
public static void main(String[] args) {
final int NUM_PHILOSOPHERS = 5;
Lock[] chopsticks = new Lock[NUM_PHILOSOPHERS];
for (int i = 0; i < NUM_PHILOSOPHERS; i++) {
chopsticks[i] = new ReentrantLock();
}
Thread[] philosophers = new Thread[NUM_PHILOSOPHERS];
for (int i = 0; i < NUM_PHILOSOPHERS; i++) {
philosophers[i] = new Thread(new Philosopher(i, chopsticks[i], chopsticks[(i + 1) % NUM_PHILOSOPHERS]));
philosophers[i].start();
}
}
}
```
在这个代码中,每个哲学家是一个线程,左边和右边的筷子是一个`Lock`对象,通过`tryLock()`方法尝试获取筷子的锁。如果成功获取了两支相邻的筷子的锁,就可以就餐一段时间;如果没有获取到右边的筷子的锁,则需要放下左边的筷子的锁,等待一段时间再重新尝试就餐。这个程序会一直运行,直到强制终止。