One liner to convert from list<T> to vector<T>
Is there an one-liner that converts a list<T>
to vector<T>
?
A google search returns me a lot of results that use manual, lengthy conversion, which make me puke. Should we go to that much trouble to do something as simple as a list-to-vector conversion?
Solution 1:
You can only create a new vector with all the elements from the list:
std::vector<T> v{ std::begin(l), std::end(l) };
where l
is a std::list<T>
. This will copy all elements from the list to the vector.
Since C++11 this can be made more efficient if you don't need the original list anymore. Instead of copying, you can move all elements into the vector:
std::vector<T> v{ std::make_move_iterator(std::begin(l)),
std::make_move_iterator(std::end(l)) };
Solution 2:
The accepted answer of:
std::vector<T> v(std::begin(l), std::end(l));
is certainly correct, but it's (quite unfortunately) not optimal given the recent change in requirement that std::list::size()
be O(1)
. If you have a conforming implementation of std::list
(which, for instance, gcc didn't have until 5+), then the following is quite a bit faster (on the order of 50% once we get to 50+ elements):
std::vector<T> v;
v.reserve(l.size());
std::copy(std::begin(l), std::end(l), std::back_inserter(v));
It's not a one liner, but you could always wrap it in one.
Solution 3:
How about this?
list<T> li;
vector<T> vi;
copy(li.begin(),li.end(),back_inserter(vi));