In this SO answer and this proposal we can see an implementation (included below for convenience) for std::is_specialization_of
, which can detect if a type is a specialization of a given template.
template< class T, template<class...> class Primary >
struct is_specialization_of : false_type;
template< template<class...> class Primary, class... Args >
struct is_specialization_of< Primary<Args...>, Primary> : true_type;
The proposal explicitly says that this type trait is not affected by inheritance:
The proposed trait considers only specialization. Since specialization is unrelated to inheritance, the trait’s result is unaffected when any template argument happens to be defined via inheritance.
template< class > struct B { };
template< class T > struct D : B<T> { };
static_assert( is_specialization_of_v< B<int>, B> );
static_assert( is_specialization_of_v< D<int>, D> );
static_assert( not is_specialization_of_v< B<int>, D> );
static_assert( not is_specialization_of_v< D<int>, B> );
Is there a way to implement something that does take inheritance into consideration, behaving like std::derived_from
but where the Base
could be any specialization of a given template? Something like, say std::derived_from_specialization_of
:
template<class> struct A {};
template<class T> struct B : A<T> {};
template<class T> struct C : B<T> {};
static_assert(derived_from_specialization_of< A<int>, A>);
static_assert(derived_from_specialization_of< B<int>, B>);
static_assert(derived_from_specialization_of< C<int>, C>);
static_assert(derived_from_specialization_of< B<int>, A>);
static_assert(derived_from_specialization_of< C<int>, B>);
static_assert(derived_from_specialization_of< C<int>, A>);
It should also support templates with multiple parameters, and/or parameter packs:
template<class, class> struct A{};
template<typename...> struct B{};
CodePudding user response:
template <template <class...> class Template, class... Args>
void derived_from_specialization_impl(const Template<Args...>&);
template <class T, template <class...> class Template>
concept derived_from_specialization_of = requires(const T& t) {
derived_from_specialization_impl<Template>(t);
};
which we use to implement the exposition-only type trait is-derived-from-view-interface
from [range.view]/6.
Note that this has the same limitation as the is_specialization_of
trait cited in the OP: it works only with templates that have all type parameters.