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.
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;
}
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.