인프런 커뮤니티 질문&답변

안녕하세요 

마지막에

	cout << "Future" << endl;
	{
		const auto sta = chrono::steady_clock::now();

		unsigned long long sum = 0;

		vector<std::future<int>> futures;
		futures.resize(n_threads);

		const unsigned n_per_thread = n_data / n_threads; // assumes remainder = 0
		for (unsigned t = 0; t < n_threads; ++t)
			futures[t] = std::async(dotProductFuture, std::ref(v0), std::ref(v1),
						t * n_per_thread, (t + 1) * n_per_thread);
		 
		for (unsigned t = 0; t < n_threads; ++t)
			sum += futures[t].get();

		const chrono::duration<double> dur = chrono::steady_clock::now() - sta;

		cout << dur.count() << endl;
		cout << sum << endl;
		cout << endl;
	}

이 코드를 쓰레드와 함께해서 사용해보고 promise 사용해보라고 하셨는데

// TODO: use divide and conquer strategy for std::thread

// TODO: use promise

	cout << "Thread future" << endl;
	{
		const auto sta = chrono::steady_clock::now();


		vector<thread> threads;
		threads.resize(n_threads);
		
		std::promise<int> p;
		auto f = p.get_future();


		const unsigned n_per_thread = n_data / n_threads; // assumes remainder = 0
		for (unsigned t = 0; t < n_threads; ++t)
			threads[t] = std::thread(dotProductEx1, std::ref(v0), std::ref(v1),
									t * n_per_thread, (t + 1) * n_per_thread, move(p));

		for (unsigned t = 0; t < n_threads; ++t)
			threads[t].join();

		const chrono::duration<double> dur = chrono::steady_clock::now() - sta;
		
		int i = f.get();

		cout << dur.count() << endl;
		cout << i << endl;
		cout << endl;
	}

void dotProductEx1(const vector<int> &v0, const vector<int> &v1,
				   const unsigned i_start, const unsigned i_end, std::promise<int> &&p)
{
	int sum = 0; // local sum
	for (unsigned i = i_start; i < i_end; ++i)
		sum += v0[i] * v1[i]; //

	p.set_value(sum);
}

이렇게 해봤는데

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