Rickie Rickie - 1 month ago 14
C++ Question

SFINAE does not work as expected

I am reading this article and find this piece of code using

SFINAE
very interesting:

template<typename D, typename T2, typename... Args>
struct get_value_int<D, T2, Args...> {
template<typename D2, typename T22, typename Enable = void>
struct impl
: std::integral_constant<int, get_value_int<D, Args...>::value> {};

template<typename D2, typename T22>
struct impl <D2, T22, std::enable_if_t<std::is_same<typename D2::type_id, typename T22::type_id>::value>>
: std::integral_constant<int, T22::value> {};

static constexpr const int value = impl<D, T2>::value;
};


As I understand, if the template class
std::enable_if_t
could not be instaniated, the first version of
impl
would be.

However, when I try to write a simpler piece of code, which I think pretty much the same as the one above, it gets compile errors. Do I misunderstand something here?

template<typename T,typename N = void>
class A
{

};

template <typename T>
class A<T, typename enable_if<false>::type>
{

};

Answer

SFINAE occurs in immediate context, here your condition doesn't depend of T so it is a hard failure, your code should be something like:

template <typename T>
class A<T, typename enable_if<condition<T>::value>::type>
{
};