CODEWITHSUNDEEP
c++templatesmetaprogramming
Fausturs
Fausturs

Reputation: 81

c++ metaprogram, member type test?

I try to write a template meta function to test "is a type T has a member type, named type".

here is the code:


#include <iostream>
#include <type_traits>

template <typename T, typename E = void>
struct has_type;
// branch 1
template <typename T, typename E>
struct has_type : std::false_type {};
// branch 2
template <typename T>
struct has_type<
    T, std::enable_if_t<std::is_same_v<typename T::type, typename T::type>, T>
> : std::true_type {};


struct with_type {using type = void;};
struct without_type {};

int main()
{
    std::cout<< has_type< with_type >::value <<std::endl;
    std::cout<< has_type< without_type >::value <<std::endl;
    return 0;
}

I imagined, compiler will first try to use branch 2, if the type T has the member type type, we get std::true_type. Or fail to find T::type, then SFINAE and use branch 1, we get std::false_type

But the two output is both false.

Is there anything I understand wrong?

Upvotes: 1

Views: 73

Answers (1)

songyuanyao
songyuanyao

Reputation: 172964

The branch 1, i.e. the primary template, the default value of the 2nd template parameter is void; to make the branch 2, i.e. the specialization to be selected, the 2nd template argument should yield type void instead of T, when the condition is satisfied.

// branch 2
template <typename T>
struct has_type<
    T, std::enable_if_t<std::is_same_v<typename T::type, typename T::type>, void>
//                                                                          ^^^^
> : std::true_type {};

Or just

template <typename T>
struct has_type<
    T, std::void_t<typename T::type>
> : std::true_type {};

LIVE

Upvotes: 4

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