C++ Thread 信号量

信号量实现生产者消费者

int g_num = 0;
std::binary_semaphore semp(1);
std::binary_semaphore sems(0);

void P(int i)
{
	for (int i = 0; i < 10; ++i)
	{
		semp.acquire(); //1 -> 0    P操作
		g_num = i;
		cout << "P:" << g_num << endl;
		sems.release(); //0 -> 1    V操作
	}
}

void S()
{
	for (int i = 0; i < 20; ++i)
	{
		sems.acquire();
		cout << "S:" << g_num << endl;
		semp.release();
	}
}

int  main()
{
	thread tha(P, 1);
	thread thc(P, 2);
	thread thb(S);

	tha.join();
	thb.join();
	thc.join();

	return 0;
}

int g_num = 0;
std::counting_semaphore semp(2);
std::counting_semaphore sems(0);
std::mutex mx;

void P(int i)
{
	for (int i = 0; i < 10; ++i)
	{
		semp.acquire(); //1 -> 0    P操作
		mx.lock();
		g_num = i;  //防止g_num线程不完全
		mx.unlock();
		cout << "P:" << g_num << endl;
		sems.release(); //0 -> 1    V操作
	}
}

void S()
{
	for (int i = 0; i < 20; ++i)
	{
		sems.acquire();
		cout << "S:" << g_num << endl;
		semp.release();
	}
}

int  main()
{
	thread tha(P, 1);
	thread thc(P, 2);
	thread thb(S);

	tha.join();
	thb.join();
	thc.join();

	return 0;
}

MySemaphore

class MySemaphore
{
public:
	MySemaphore(int val = 1):count(val){}
	void P()  //acquire()
	{	
		std::unique_lock<std::mutex> lck(mtk);
		if (--count < 0) //资源不足
		{
			cv.wait(lck); //一旦唤醒则代表有资源
		}
	}
	void V()  //release()
	{
		std::unique_lock<std::mutex> lck(mtk);
		if (++count <= 0)
		{ 
			cv.notify_one();
		}
	}
	int get_count()
	{
		return count;
	}

private:
	int count;
	std::mutex mtk;
	std::condition_variable cv;
};

int g_num = 0;
MySemaphore semp(1);
MySemaphore sems(0);

std::mutex mx;
void P(int id)
{
	for (int i = 0; i < 10; ++i)
	{
		semp.P();  // p =  1 -> 0
		mx.lock();
		g_num = i;
		cout << "生产者：" << "  " << g_num << endl;
		mx.unlock();
		sems.V();  // s = 0 -> 1
	}
}
void S(int id)
{
	for (int i = 0; i < 10; ++i)
	{
		sems.P();
		mx.lock();
		cout << "消费者: " << ":" << g_num << endl;
		mx.unlock();
		semp.V();

	}
}

int main()
{
	std::thread tha(P, 0);
	std::thread ths(S, 0);

	tha.join();
	ths.join();

	return 0;
}

1. 首先生产者线程进入，对生产者的信号量进行了一次P操作，生产者的信号量Pcount 1 -> 0 ，并且在最后将消费者的信号量Scount由0 -> 1
2. 紧接消费者进入线程，进行一次P操作，消费者信号量Scount 1 -> 0，然后生产者的信号量进行一次V操作 Pcount 0 -> 1
3. 倘若下一次进入的还是消费者，首先进行一次P操作，那么消费者Scount 0 -> -1 ，则消费者P操作进入等待队列
4. 这时消费者的P操作陷入等待队列，只能由生产者进行P操作，则生产者信号量自减 Pcount 1 -> 0，接着将消费者进行V操作 Scount -1 -> 0 ，此时进行一次唤醒，将原本进入等待队列的消费者唤醒