Move boost :: ptr_vector

Given a type that is not copied but moves, I am trying to build boost::ptr_vector

that type and return it from a function. I am using Visual Studio 2013.

struct NonCopyableButMovable
{
    using this_type = NonCopyableButMovable;

    this_type() {}

    this_type(this_type const &) = delete;
    this_type & operator =(this_type const &) = delete;

    this_type(this_type &&) {} // no default move ctr in VS2013
    this_type & operator =(this_type &&) {} // no default move assignment in VS2013
};

boost::ptr_vector<NonCopyableButMovable> make_ptr_vector()
{
    boost::ptr_vector<NonCopyableButMovable> res;
    //return res; // (1)
    //return std::move(res); // (2)
    return boost::ptr_vector<NonCopyableButMovable>(); // (3)
}

      

        
        
        
      

    

I can only get (3) to compile: the temporary is ptr_vector

moved. With (1) and (2), the compiler calls boost::new_clone

, which tries to call the NonCopyableButMovable

copy constructor .

According to this FAQ , the "Moving out of functions" section, (1) should work, ie. res

should be moved, not copied.

According to this thread , (2) can be used as a workaround for compilers that are not fully standard-compliant, but for some reason I don't understand is boost::new_clone

also called with (2).

Note that if I replace boost::ptr_vector

with std::vector

, the three return methods are compiled.

Is there something broken in the Boost ptr_containers move semantics with VS2013? What happens in case (2)? What would be the workaround to implement such a ptr_container factory function?