(과제스포)제가 과제를 똑바로 이해했는지 궁금합니다.

#include<chrono>
#include<iostream>
#include<mutex>
#include<random>
#include<thread>
#include<utility>
#include<vector>
#include<atomic>
#include<future>
#include<numeric>
#include<execution>

using namespace std;

mutex mtx;


void dotProductDQThread(const vector<int>& v0, const vector<int>& v1,
	const unsigned i_start, const unsigned i_end, unsigned long long& sum)
{
	int sum_tmp = 0; //local sum
	for (unsigned i = i_start; i < i_end; ++i)
	{
		sum_tmp += v0[i] * v1[i];
	}
	sum += sum_tmp;
}

void dotProductProm(const vector<int>& v0, const vector<int>& v1,
	const unsigned i_start, const unsigned i_end, promise<unsigned long long>&& sum)
{
	int sum_tmp = 0; //local sum
	for (unsigned i = i_start; i < i_end; ++i)
	{
		sum_tmp += v0[i] * v1[i];
	}
	sum.set_value(sum_tmp);
}



int main()
{
	/*v0 = { 1,2,3 };
	v1 = { 4,5,6 };
	v0_dot = 1 * 4 + 2 * 5 + 3 * 6;*/

	const long long n_data = 100'000'000;
	const unsigned n_threads = 4;

	//initialize vectors
	std::vector<int> v0, v1;
	v0.reserve(n_data);
	v1.reserve(n_data);

	random_device seed;
	mt19937 engine(seed());

	uniform_int_distribution<> uniformDist(1, 10);

	//v0와 v1에 값 무작위로 넣어줌
	for (long long i = 0; i < n_data; ++i)
	{
		v0.push_back(uniformDist(engine));
		v1.push_back(uniformDist(engine));
		//cout << v0[i] << "\t" << v1[i] << endl;
	}


	cout << "std::inner_product" << endl;
	{
		const auto sta = chrono::steady_clock::now();
		const auto sum = std::inner_product(v0.begin(), v0.end(), v1.begin(), 0ull);//0ull = unsigned longlong 0
		const chrono::duration<double> dur = chrono::steady_clock::now() - sta;

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


	//TODO: use divde and conquer strategy for std::thread
	cout << "TODO: use divde and conquer strategy for std::thread" << endl;
	{
		const auto sta = chrono::steady_clock::now();
		unsigned long long sum = 0;

		vector<thread> threads;
		threads.resize(n_threads);

		const unsigned n_per_thread = n_data / n_threads;
		for (unsigned t = 0; t < n_threads; ++t)
			threads[t] = std::thread(dotProductDQThread, std::ref(v0), std::ref(v1),
				t * n_per_thread, (t + 1) * n_per_thread, std::ref(sum));

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

		const chrono::duration<double> dur = chrono::steady_clock::now() - sta;
		cout << dur.count() << endl;
		cout << sum << endl;
		cout << endl;
	}

	//TODO: use promise
	cout << "TODO: use promise" << endl;
	{
		const auto sta = chrono::steady_clock::now();
		//std::promise<unsigned long long> sum;
		
		auto sum = 0ull;
		vector<thread> threads;
		vector<future<unsigned long long>> futures;
		vector<promise<unsigned long long>> proms;
		threads.resize(n_threads);
		futures.resize(n_threads);
		proms.resize(n_threads);

		const unsigned n_per_thread = n_data / n_threads;
		for (unsigned t = 0; t < n_threads; ++t)
		{
			futures[t] = proms[t].get_future();
			threads[t] = std::thread(dotProductProm, std::ref(v0), std::ref(v1),
				t * n_per_thread, (t + 1) * n_per_thread, std::move(proms[t]));
		}

		for (unsigned t = 0; t < n_threads; ++t)
		{
			threads[t].join();
			sum += futures[t].get();
		}


		const chrono::duration<double> dur = chrono::steady_clock::now() - sta;
		cout << dur.count() << endl;
		cout << sum << endl;
		cout << endl;
	}

}

결과는 제대로 나오지만, 제가 과제를 똑바로 이해했는지 궁금해서 여쭤봅니다.

1. 과제1: use divde and conquer strategy for std::thread
   쓰레드에 sum에 local sum값을 넣어 race condition 해결

2. 과제2: prom 사용해보기
   sum 변수 선언 및 thread, promise, future 모두 쓰레드 크기 만큼의 vector로 만들어줬습니다.
   병렬 처리 후 future값을 sum에 더해줬습니다.
   
   ※추가 궁금증
   promise 과제 중, std::ref와 std::move 둘 다 해보았습니다. 두 경우 모두 정상 작동하였는데, 어떤 방법을 가장 추천하시나요?