How to get an element from std :: tuple by type

Question

I have a set of classes A, B, C and I want to have access instances from a common code by type, fe

template<typename T>
newObject()
{
    return m_storage->getNew();
}

where m_storage is an instance of A or B or C, depends on T.

So, I came up with std :: tuple, but there is a problem because I cannot get the element from the tuple by type.

std::tuple<A,B,C> m_tpl;
template<typename T>
newObject()
{
    return m_tpl.get<T>().getNew();
}

Is there any way to do this? Is it possible?

Thanks.

PS: I do not want to write a newObject specialization for each type. :-)

Answer 1

This is a project from C ++ 14 about getting a value from a tuple by type.

But before C ++ 14 arrives, you can write something like below:

namespace detail
{

template <class T, std::size_t N, class... Args>
struct get_number_of_element_from_tuple_by_type_impl
{
    static constexpr auto value = N;
};

template <class T, std::size_t N, class... Args>
struct get_number_of_element_from_tuple_by_type_impl<T, N, T, Args...>
{
    static constexpr auto value = N;
};

template <class T, std::size_t N, class U, class... Args>
struct get_number_of_element_from_tuple_by_type_impl<T, N, U, Args...>
{
    static constexpr auto value = get_number_of_element_from_tuple_by_type_impl<T, N + 1, Args...>::value;
};

} // namespace detail

template <class T, class... Args>
T get_element_by_type(const std::tuple<Args...>& t)
{
    return std::get<detail::get_number_of_element_from_tuple_by_type_impl<T, 0, Args...>::value>(t);
}

int main()
{
    int a = 42;

    auto t = std::make_tuple(3.14, "Hey!", std::ref(a));

    get_element_by_type<int&>(t) = 43;

    std::cout << a << std::endl;

    // get_element_by_type<char>(t); // tuple_element index out of range

    return 0;
}

Answer 2

A simple mixin variation container does the trick:

template < typename T > struct type_tuple_value 
{ 
    T value; 
    type_tuple_value ( T&& arg ) : value(std::forward<T>(arg)) {}
};

template < typename ...T > struct type_tuple : type_tuple_value<T>...
{
    template < typename ...Args > type_tuple ( Args&&... args ) :
    type_tuple_value<T>(std::forward<T>(args))... {}
    template < typename U > U& get() { return type_tuple_value<U>::value; }
    template < typename U > const U& get() const { return type_tuple_value<U>::value; }
};

Example

Answer 3

You can also calculate the type position using the constexpr function if you don't like the template.

  constexpr int count_first_falses() { return 0; }

  template <typename... B>
  constexpr int count_first_falses(bool b1, B... b)
  {
    if (b1) return 0;
    else return 1 + count_first_falses(b...);
  }

  template <typename E, typename... T>
  decltype(auto) tuple_get_by_type(const std::tuple<T...>& tuple)    
  {
    return std::get<count_first_falses((std::is_same<T, E>::value)...)>(tuple);
  }