Template alias for type trait doesn't work

Question

I'm practicing SFINAE by a simple custom type trait:

#include <iostream>
#include <type_traits>

struct A{ int i, j; };

// Type trait
template<typename T>
struct is_class_A{ static const bool value = false; };

template<>
struct is_class_A<A>{ static const bool value = true; };

// (1)
template <typename T>
std::enable_if_t<is_class_A<T>::value> foo(const T& t){
    puts("Hi!\n");
}


int main() {
    A a;
    foo(a);   // allowed
    //foo(1); // not allowed
}

works as expected. However, using a template alias

template <typename T>
using is_class_A_v = is_class_A<T>::value;


// (1)
template <typename T>
std::enable_if_t<is_class_A_v<T>> foo(const T& t){
    puts("Hi!\n");
}

yields the compiler error:

<source>:37:22: error: missing 'typename' prior to dependent type name 'is_class_A<T>::value'
using is_class_A_v = is_class_A<T>::value;
                     ^~~~~~~~~~~~~~~~~~~~
                     typename 
<source>:40:18: error: template argument for non-type template parameter must be an expression
std::enable_if_t<is_class_A_v<T>> foo(const T& t){

On the other hand, by using

template <typename T>
using is_class_A_v = typename is_class_A<T>::value;

I obtain the error message

error: template argument for non-type template parameter must be an expression
std::enable_if_t<is_class_A_v<T>> foo(const T& t){
                 ^~~~~~~~~~~~~~~

What am I missing here?

Answer 1

Template variable syntax would be:

template <typename T> const bool is_class_A_v = is_class_A<T>::value;