用c++编程实现哲学家进餐问题

好的，以下是用C++编程实现哲学家进餐问题的代码：

#include <iostream>
#include <thread>
#include <mutex>

using namespace std;

const int kNumPhilosophers = 5;
mutex forks[kNumPhilosophers];

void philosopher(int id) {
    int left_fork = id;
    int right_fork = (id + 1) % kNumPhilosophers;

    while (true) {
        // philosopher is thinking
        cout << "Philosopher " << id << " is thinking." << endl;

        // philosopher is hungry and tries to acquire forks
        forks[left_fork].lock();
        forks[right_fork].lock();

        // philosopher has both forks and can eat
        cout << "Philosopher " << id << " is eating." << endl;

        // philosopher is done eating and releases forks
        forks[right_fork].unlock();
        forks[left_fork].unlock();
    }
}

int main() {
    thread philosophers[kNumPhilosophers];

    for (int i = 0; i < kNumPhilosophers; i++) {
        philosophers[i] = thread(philosopher, i);
    }

    for (int i = 0; i < kNumPhilosophers; i++) {
        philosophers[i].join();
    }

    return 0;
}

该程序模拟了5个哲学家同时进餐的情况。每个哲学家都有一个左手和右手的叉子，每次进餐时需要同时获得这两个叉子才能开始进餐。为了防止死锁，实现中使用了"避免等待"策略，即每个哲学家先尝试获取左边的叉子，再尝试获取右边的叉子。如果获取不到，就放下已经获取的叉子重新尝试获取。如果每个哲学家都遵守这个策略，就可以避免死锁的情况。