Moving a vector element to the back of the vector

Question

Is there any better way (either faster or with fewer symbols of code) than erasing the element and re-adding it to the back?

template 
void mov         


        

Answer 1

You can avoid the extra variable.

v.push_back(v[itemIndex]);
v.erase(v.begin() + itemIndex);

If you delete frequently from the middle of the vector and can rewrite your code so that it doesn't require random access, you may be able to improve efficiency by using a linked list (std::list) instead.

Answer 2

You can do this with std::rotate from the standard library. Since this doesn't change the vector size it also won't trigger a reallocation. Your function would look something like this:

template <typename T>
void moveItemToBack(std::vector<T>& v, size_t itemIndex)
{
    auto it = v.begin() + itemIndex;
    std::rotate(it, it + 1, v.end());
}

Answer 3

Possibly the fastest way, would be to swap it with the last element

template <typename T>
void moveItemToBack(std::vector<T>& v, size_t itemIndex)
{
   std::swap(v[itemIndex], v.back()); // or swap with *(v.end()-1)
}

one operation! Ofcourse std::swap has to work with T