将std :: vector移动到C ++ 11中的std :: deque

Question

If I have std::deque and std::vector and want to combine them to std::deque , I can do it the following way:

typedef int T; // type int will serve just for illustration
std::deque< T > deq(100); // just some random size here
std::vector< T > vec(50);
// ... doing some filling ...
// now moving vector to the end of queue:
deq.insert( 
    deq.end(), 
    std::make_move_iterator( vec.begin() ),
    std::make_move_iterator( vec.end() )
);
std::cout << deq.size() << std::endl;

We know the size of the vector but we can't reserve memory at the end of std::deque before using std::deque.insert(...) . So is it the fastest way to move all elements of std::vector to the end of std::deque ? Or did I miss something?

Thank you.

Answer 1

I would use resize method as follows, because than is deque reallocated only once:

size_t oldSize = deq.size();
deq.resize(deq.size() + vec.size());
copy(vec.begin(), vec.end(), deq.begin() + oldSize);

Answer 2

try this:

using T = int; // type int will serve just for illustration

std::deque< T > deq(100); // just some random size
std::vector< T > vec(50);
// ... doing some filling ...
// now moving vector to the end of queue:
std::move( 
    begin(vec),
    end(vec),
    back_inserter(deq)
);
std::cout << deq.size() << std::endl;

Keep in mind that this still copies the vector to the end of the deq . It just applies std::move on each element of vec to the end of deq . As long as T is just an int this is not much different than copying the vector to the end of deq .