C++11多线程模拟售票

骑猪看日落 2024-04-18 12:44 30阅读 0赞

买火车票是大家春节回家最为关注的事情，我们就简单模拟一下火车票的售票系统

(为使程序简单，我们就抽出最简单的模型进行模拟)：有100张从北京到赣州的火车票，在6个窗口同时出售，保证系统的稳定性和数据的原子性。）

#include <iostream>
    #include <list>
    #include <thread>
    #include <mutex>
    #include <chrono>
    #include <future>
    #include <condition_variable>
    std::condition_variable cv;
    std::mutex mutx;
    int ticket = 100;
    bool isrunning = true;
    void consumer(int id)
    {
        
    	while (ticket > 0)
    	{
        
    		std::this_thread::sleep_for(std::chrono::milliseconds(100));
    		std::lock_guard<std::mutex> lck(mutx);
    		if(ticket > 0)
    		{
        
    			int x = (rand() % (10 - 1)) + 1;
    			if (x > ticket)
    				x = ticket;
    			std::cout << "售票窗口[" << id << "]: ";
    			ticket -= x;
    			std::cout << "售出 " << x << "票" << std::endl;
    			std::cout << "剩余 " << ticket << "票" << std::endl;
    		}
    		else
    		{
        
    			std::cout << "售票窗口[" << id << "]: ";
    			std::cout << "票已经售完..." << std::endl;
    		}
    	}
    }
    int main()
    {
        
    	std::cout << "main thread:" << std::this_thread::get_id() << std::endl;
    	std::cout << "总共票数为"  << ticket << std::endl;
    	std::thread c[5];
    	for (int i = 0; i < 5; i++)
    		c[i] = std::thread(consumer, i+1);
    	for (int i = 0; i < 5; i++)
    		c[i].join();
    	std::cout << "it's over..." << std::endl;
    	return 0;
    }

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