Moving array of unique_ptr<T> in a recursive data structure

Question

Attempting to compile the following code results in the following compile error:

error C2280: 'std::unique_ptr>::unique_ptr(const std::unique_ptr<_Ty,std::default_delete<_Ty>> &)' : attempting to reference a deleted function

My understanding is that the array 'm_children' should be movable since the type pointed to by unique_ptr has a move constructor defined.

Unless this is an error being caused by the recursive nature of the class or some element of move semantics that I have overlooked?

#include <array>
#include <memory>
#include <iostream>

class OctreeNode{
public:
    OctreeNode(){ };
    OctreeNode(OctreeNode&& other) : m_children(std::move(other.m_children)){};
private:
    std::array<std::unique_ptr<OctreeNode>, 8> m_children;
};

int main(int argc, char* argv[])
{
    OctreeNode T;
    std::cout << "Success!" << std::endl;
    return 0;
}

Answer 1

It seems like in vs2013 that definition of std::array in the <array> header doesn't include a move constructor or move assignment operator.

Under C++ rules these should be automatically generated (and so unnecessary to manually define), but msvc didn't include implicit generation of those until vs2015.

Answer 2

according to the error message, it doesn't call the move constructor of the unique_ptr. Instead, it calls the copy-constructor of it which is not exist. So, iterate over the unique_ptrs and call move constructor explicitly.

    int i = 0;
    for(auto& x : other.m_children)
        m_children[i++] = std::move(x);

Btw, normally it should work because std::array is movable if and only if the element that it stores is movable too. In you case, unique_ptr is movable so that it shouldn't be problem. Please refer to below link for more detailed discussion

Is std::array movable?