0
template
<
    template <typename, typename>
        class storage_t,
    typename T,
    typename is_allocated
>
class Buffer : public storage_t<T, is_allocated> { ... };

template
<
    template <typename, typename>
        class storage_t,
    typename T = storage::UnknownType,
    typename is_allocated = std::false_type
>
class Example_Buffer
: public Buffer<storage_t, T, is_allocated> {
    constexpr Example_Buffer(
        typename storage_t<T, is_allocated>::iterator it) {}
};

Example_Buffer<...> inherits from Buffer<...>. Buffer<storage_t, T, is_allocated> inherits from storage_t<T, is_allocated>. storage_t<...> includes typedefs and static constexpr data. Is there a way to access these typedefs and static constexpr data in the constructor of Example_Buffer through inheritance? (through inheritance, that is, not using storage_t<T, is_allocated>? It feels slightly odd to use this syntax twice within the same class.

Feel free to ask if I need to elaborate.

Lourens Dijkstra
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1 Answers1

1

The members are inherited and accessible as long as they are public in storage_t. You can use the injected class name and/or the injected base class name to access these dependent members,

Here are a few options..

#include <iostream>
#include <type_traits>
#include <typeinfo>

using namespace std;

template <typename A,typename B>
struct basic_storage
{
    using a_type = A;
    using b_type = B;
    static constexpr bool value = b_type::value;
};


template
<
    template <typename, typename>
        class storage_t,
    typename T,
    typename is_allocated
>
class Buffer : public storage_t<T, is_allocated> {

    public:
    using storage_type = storage_t<T, is_allocated>;
};

template
<
    template <typename, typename>
        class storage_t,
    typename T /*= storage::UnknownType*/,
    typename is_allocated = std::false_type
>
class Example_Buffer
: public Buffer<storage_t, T, is_allocated> {
    public:
    constexpr Example_Buffer(
        /*typename storage_t<T, is_allocated>::iterator it*/) {

            //using the members with the injected class name..
            using b_type = typename Example_Buffer::b_type;

            // or directly using injected class name..
            std::cout << typeid(typename Example_Buffer::a_type).name() << std::endl;
            std::cout << Example_Buffer::value << std::endl;

            // using storage_type defined in Buffer<...>
            using storage_type = typename Example_Buffer::storage_type;

            std::cout << typeid(typename storage_type::a_type).name() << std::endl;
            std::cout << storage_type::b_type::value << std::endl;
            std::cout << storage_type::value << std::endl;

        }
};


int main() {
    Example_Buffer<basic_storage,int,std::true_type>{};
    return 0;
}

Demo

If your wondering why you cant just access then in the derived class without prefixing them with Example_Buffer:: base class name, as this post explains it is because they are dependent names.

rmawatson
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