#include <iostream>
/**** I am confused to apply sfinae method here ******/
template <typename T>
struct hasTypeFoo {
//..
static constexpr bool value = true;
};
/////////////////////////////////////////////////
struct A {
using Foo = int;
};
struct B {
};
int main()
{
constexpr bool b1 = hasTypeFoo<A>::value;
constexpr bool b2 = hasTypeFoo<B>::value;
std::cout << b1 << b2;
}
Use std::void_t:
template<typename T, typename = void>
struct hasTypeFoo : std::false_type { };
template<typename T>
struct hasTypeFoo<T, std::void_t<typename T::Foo>> : std::true_type { };
A very good explanation of how std::void_t works, can be found in this question. It is used here to silently reject the specialization if typename T::Foo is ill-formed.
You can do it with partial specialization. e.g.
// primary template
template <typename T, typename = void>
struct hasTypeFoo {
static constexpr bool value = false;
};
// partial specialization for types containing type Foo
template <typename T>
struct hasTypeFoo<T, std::void_t<typename T::Foo>> {
static constexpr bool value = true;
};
