C++实现多线程擦桌子椅子_清扫教室c++-优快云博客

本文链接：https://blog.youkuaiyun.com/Zoe_Cpp/article/details/125982386

本文介绍了一个使用C++实现的多线程程序，通过`Work`类，两个线程轮流清理桌子和椅子，展示了互斥锁在并发控制中的应用。工作量统计显示了每个线程的擦桌擦椅次数。

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#include <iostream>
#include <functional>
#include <thread>
#include <mutex>
#include <unordered_map>
#include <cstdlib>
class Work {
private:
	//锁，防止多个线程同时修改或访问同一个数据
	std::mutex deskMutex;
	std::mutex chairMutex;
	//桌子总数
	int deskCount = 500;
	//椅子总数
	int chairCount = 500;
public:
	//存贮每个线程擦桌子和擦椅子总数
	std::unordered_map<unsigned int, int> chairTask;
	std::unordered_map<unsigned int, int> deskTask;
	void clearDesk() {
		//上锁防止多线程同时清理同个桌子
		deskMutex.lock();
		if (deskCount > 0) {
			std::thread::id idCurrect =  std::this_thread::get_id();
			printf("线程%d\t清理了第%d个桌子\n", idCurrect, deskCount);
			deskCount--;
			//hack获取当前线程id的整数作为key，统计存储每个线程擦桌子的总数
			deskTask[*(unsigned int*)&idCurrect]++;
		}
		deskMutex.unlock();
	}
	void clearChair() {
		chairMutex.lock();
		if (chairCount > 0) {
			std::thread::id idCurrect = std::this_thread::get_id();
			printf("线程%d\t清理了第%d个椅子\n", idCurrect, chairCount);
			chairCount--;
			chairTask[*(unsigned int*)&idCurrect]++;
		}
		chairMutex.unlock();
	}
	//线程运行入口函数
	void work() {
		while (true) {
			int kase = rand() % 2;
			if (kase == 0) {
				clearDesk();
			}
			else {
				clearChair();
			}
			if (chairCount <= 0 && deskCount <= 0)
				break;
		}
	}
};
int main() {
	Work work;
	//创建两个线程并运行
	std::thread thread1(std::bind(&Work::work, &work));
	std::thread thread2(std::bind(&Work::work, &work));
	thread1.join();
	thread2.join();
	//任务结束显示每个线程的工作量
	for (auto&[name, cnt] : work.chairTask) {
		std::cout << name << "清理了" << cnt << "个椅子" << std::endl;
	}
	for (auto& [name, cnt] : work.deskTask) {
		std::cout << name << "清理了" << cnt << "个桌子" << std::endl;
	}
	return 0;
}