[C++知识库]多线程实现生产者-消费者问题——C++版本

问题描述：
1）生产者进程每次生产1个产品（产品数量大于4时停止生产）；
2）消费者进程每次消耗2个产品；

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
using namespace std;
std::mutex mu;
std::condition_variable cond;

void ProduceThread(int *productNum)
{
	while (true) {
		std::unique_lock<std::mutex> lock(mu);
		if (*productNum >= 4) {
			cout << "store is full cannot produce more." << endl;
			cond.wait(lock);
		}
		cout << "producing...." << endl;
		*productNum += 1;
		std::this_thread::sleep_for(std::chrono::milliseconds(500));
		cout << "A product was produced. current productNum:" << *productNum << endl;
		std::this_thread::sleep_for(std::chrono::seconds(2));
		cond.notify_one();
		lock.unlock();
	}
}

void ConsumeThread(int *productNum)
{
	while (true) {
		std::unique_lock<std::mutex> lock(mu);
		if (*productNum <= 1) {
			cout << "product is not enough" << endl;
			cond.wait(lock);
		}

		cout << "consuming...." << endl;
		*productNum -= 2;
		std::this_thread::sleep_for(std::chrono::milliseconds(250));
		cout << "2 products consumed." << endl;
		std::this_thread::sleep_for(std::chrono::seconds(2));
		cond.notify_one();
		lock.unlock();
	}
}
int main()
{
	int products = 3;
	std::thread proThread(ProduceThread, &products);
	std::thread conThread(ConsumeThread, &products);

	proThread.join();
	conThread.join();
	while (true) {}
	return 0;
}