std::enable_if using its internal type and without using it

Question

Why does it print "B"

#include <iostream>

template<typename T, typename U = void>
struct Test
{ static void apply() { std::cout << "A" << std::endl; } };

template<typename T>
struct Test<T, typename std::enable_if<true>::type>
{ static void apply() { std::cout << "B" << std::endl; } };

int main()
{
    Test<int>::apply();
}

but this one "A"?:

#include <iostream>

template<typename T, typename U = void>
struct Test
{ static void apply() { std::cout << "A" << std::endl; } };

template<typename T>
struct Test<T, std::enable_if<true>>
{ static void apply() { std::cout << "B" << std::endl; } };

int main()
{
    Test<int>::apply();
}

The only difference between them is that, in the first one, I'm using typename std::enable_it<true>::type as U (e.g, void), but in the second one, I'm directly using std::enable_if<true> as U, which is also a well defined type and with no more meaning as void.

Answer 1

Because when you write Test<int> you are actually writing Test<int, void> which will never be the same as Test<int, std::enable_if<true>> (though it is the same as Test<int, typename std::enable_if<true>::type>)