I wonder how to conditionally set a type alias, based on the existance of a type alias in input argument like this.
struct a { using type = int; }
template <typename T> struct wrapper {
using innter_t = ???how???; // if T::type exists, use T::type, else T
};
static_assert(std::is_same<wrapper<int>::innter_t, int>, "expected int");
static_assert(std::is_same<wrapper<a>::innter_t, int>, "expected inner type: int");
One naive attempt is like std::conditional<std::is_class<T>, T::type, T>
if the existence of a type
alias is a prerequisite, but is far from being safe, lacking a check for something like is_alias_present<T, type>
. Besides that T::type
is potentially not existing for all types resulting in a compiler error.
Apearently std::experimental::is_detected looks promising, but unfortunately not an option for now.
Not sure if metaprogramming libraries like boost::mp11
or boost::hana
are a good approach to get that done.
CodePudding user response:
With the help of C 20 concepts
template<class T>
struct innter {
using type = T;
};
template<class T>
requires requires { typename T::type; }
struct innter<T> {
using type = T::type;
};
template <typename T>
struct wrapper {
using innter_t = innter<T>::type; // if T::type exists, use T::type, else T
};
In C 17, you can use void_t
to detect the validity of T::type
#include <type_traits>
template<class T, class = void>
struct innter {
using type = T;
};
template<class T>
struct innter<T, std::void_t<typename T::type>> {
using type = typename T::type;
};
template <typename T>
struct wrapper {
using innter_t = typename innter<T>::type; // if T::type exists, use T::type, else T
};
CodePudding user response:
One option is to write a type trait like
template <typename T, typename = void >
struct type_member
{
using type = T;
};
template <typename T >
struct type_member<T, std::void_t<typename T::type>>
{
using type = typename T::type;
};
template <typename T >
using type_member_t = type_member<T>;
and then you would use it in wrapper like
template <typename T>
struct wrapper
{
using type = type_member_t<T>::type;
};
int main()
{
static_assert(std::is_same_v<typename wrapper<int>::type, int>, "expected int");
static_assert(std::is_same_v<typename wrapper<a>::type, int>, "expected inner type: int");
}
This works by leveraging SFINAE and if typename T::type
is ill-formed in std::void_t<typename T::type>
then that specialization of type_member
is discarded and we fall back to the non-type member specialization.