How to initialize an std::array<T, 2> where T is non-copyable and non-default-constructible?

Question

I have a fixed number of objects of class T that are non-copyable and non-default-constructible. Since the size is fixed I would like to use an array-ish container like std::array instead of unique_ptr or vector . I would like to avoid the additional layer of indirection if I can help it.

How do I initialize an std::array<T, 2> ? Using array<T, 2> {T(...), T(...)} results in an error about deleted copy constructor. Using array<T, 2> {move(T(...)), move(T(...))} does not force the array elements to use the move constructor. If std::array<T, 2> inherently does not work, what else can I do without resorting to an additional layer of indirection or manual memory management techniques like placement-new?

Answer 1

No need for extra stuff, just initialize it directly:

class Foo {
public:
    Foo() = delete;
    Foo(int,char) {}

    Foo(Foo const &) = delete;
    Foo & operator = (Foo const &) = delete;
};

    std::array<Foo, 2> arr {{ {1, '1'}, {2, '2'} }};

DEMO

Answer 2

My example is not the best but it works.. Just create a function that make_array and have it create the instances for you..

http://ideone.com/fxAO3t

#include <array>
#include <iostream>

class Foo
{
    public:
        Foo(int)
        {

        }

        Foo(const Foo &) = delete;

        Foo(Foo &&)
        {
            std::cout<<"Moved\n";
        }
};

template<class... Type>
constexpr std::array<typename std::common_type<Type...>::type, sizeof...(Type)> make_array(Type&&... t)
{
    return {std::forward<Type>(t)...};
}

int main() {
    auto arr = make_array(Foo{1}, Foo{2});
    return 0;
}

As pointed out in the comments, you can do std::array<Foo, 2> arr = {Foo{1}, Foo{2}} .

I noticed you were trying to move in your post. Are you sure your class is moveable ?