java哲学家进餐问题
时间: 2023-10-29 09:51:05 浏览: 51
Java哲学家进餐问题是一个经典的并发问题,描述了多个哲学家在一张圆桌上进餐的情景。这些哲学家围坐在一张圆桌周围,每个哲学家面前有一盘饭和一只叉子。每个哲学家思考和进食的时间不一定相同,他们需要通过拿起左右两只叉子来进食,但是叉子是共享资源,只有同时拿到左右两只叉子才能进食。如果两个相邻的哲学家同时拿起了相邻的叉子,那么就会发生死锁。为了避免死锁,需要实现一个合理的进餐策略,例如只允许最多4个哲学家同时进餐,或者通过随机等待一段时间再尝试拿起叉子等方式来避免死锁。
相关问题
java 哲学家进餐问题
哲学家进餐问题是一个经典的多线程同步问题,Java中可以使用多种方法来解决这个问题,以下是其中两种方法:
方法一:使用ReentrantLock和Condition
```java
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.ReentrantLock;
public class DiningPhilosophers {
private ReentrantLock[] forks;
private Condition[] conditions;
public DiningPhilosophers(int num) {
forks = new ReentrantLock[num];
conditions = new Condition[num];
for (int i = 0; i < num; i++) {
forks[i] = new ReentrantLock();
conditions[i] = forks[i].newCondition();
}
}
public void startEating(int i) throws InterruptedException {
int leftFork = i;
int rightFork = (i + 1) % forks.length;
if (i % 2 == 0) {
forks[leftFork].lock();
forks[rightFork].lock();
} else {
forks[rightFork].lock();
forks[leftFork].lock();
}
System.out.println("Philosopher " + i + " is eating.");
Thread.sleep(1000);
forks[leftFork].unlock();
forks[rightFork].unlock();
System.out.println("Philosopher " + i + " finished eating.");
}
}
```
方法二:使用Semaphore
```java
import java.util.concurrent.Semaphore;
public class DiningPhilosophers {
private Semaphore[] forks;
private Semaphore maxPhilosophers;
public DiningPhilosophers(int num) {
forks = new Semaphore[num];
for (int i = 0; i < num; i++) {
forks[i] = new Semaphore(1);
}
maxPhilosophers = new Semaphore(num - 1);
}
public void startEating(int i) throws InterruptedException {
int leftFork = i;
int rightFork = (i + 1) % forks.length;
maxPhilosophers.acquire();
forks[leftFork].acquire();
forks[rightFork].acquire();
System.out.println("Philosopher " + i + " is eating.");
Thread.sleep(1000);
forks[leftFork].release();
forks[rightFork].release();
maxPhilosophers.release();
System.out.println("Philosopher " + i + " finished eating.");
}
}
```
哲学家进餐问题java
哲学家进餐问题是一个经典的并发编程问题,描述了五个哲学家围坐在一张圆桌前进餐的场景,每个哲学家需要先拿起自己左右两边的筷子才能进餐,但是每次只有一根筷子可以被拿起,如果所有哲学家都拿起了自己左边的筷子,那么他们就会陷入死锁状态。解决这个问题的方法有很多,其中比较常见的是使用资源分级的方法,即让哲学家按照一定顺序拿起筷子,避免死锁的发生。
以下是一个简单的 Java 实现:
```java
public class Philosopher implements Runnable {
private int id;
private Object leftChopstick;
private Object rightChopstick;
public Philosopher(int id, Object leftChopstick, Object rightChopstick) {
this.id = id;
this.leftChopstick = leftChopstick;
this.rightChopstick = rightChopstick;
}
public void run() {
try {
while (true) {
// 拿起左边的筷子
synchronized (leftChopstick) {
System.out.println("Philosopher " + id + " picks up left chopstick");
// 拿起右边的筷子
synchronized (rightChopstick) {
System.out.println("Philosopher " + id + " picks up right chopstick and starts eating");
// 进餐
Thread.sleep(1000);
System.out.println("Philosopher " + id + " puts down right chopstick");
}
}
System.out.println("Philosopher " + id + " puts down left chopstick and thinks");
// 思考
Thread.sleep(1000);
}
} catch (InterruptedException e) {
e.printStackTrace();
}
}
public static void main(String[] args) {
int numPhilosophers = 5;
Philosopher[] philosophers = new Philosopher[numPhilosophers];
Object[] chopsticks = new Object[numPhilosophers];
for (int i = 0; i < numPhilosophers; i++) {
chopsticks[i] = new Object();
}
for (int i = 0; i < numPhilosophers; i++) {
philosophers[i] = new Philosopher(i, chopsticks[i], chopsticks[(i + 1) % numPhilosophers]);
new Thread(philosophers[i]).start();
}
}
}
```