Using SFINAE partial specialization without touching the main template

Question

I am trying to specialize a structure template for several types at once SFINAE. I know that something like the following works:

#include <iostream>

template <typename T, typename Enable = void>
struct S {
    void operator()() {
        std::cout << "Instantiated generic case" << std::endl;
    }
};

template<typename T>
using enabled_type = typename std::enable_if<
                         std::is_same<T, int>::value ||
                         std::is_same<T, float>::value
                     >::type;

template <typename T>
struct S<T, enabled_type<T>> {
    void operator()() {
        std::cout << "Instantiated int/float case" << std::endl;
    }
};

int main() {
    S<float>()();
    return 0;
}

My problem is that I cannot change the main S struct template to add typename Enable = voidas it is part of the outer header of the library. Therefore, the main template should look like this:

template <typename T>
struct S {
    void operator()() {
        std::cout << "Instantiated generic case" << std::endl;
    }
};

Can SFINAE be used to specialize this template?

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Answer 1

: , , . , void enable_if, . , :

template <class T>
struct foo {};

template <class T>
struct S<foo<T>> {};

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template <class T, class = void>
struct S {};

template <>
struct S<double> {};

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Answer 2

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struct S_int_or_float {
    void operator()() {
        std::cout << "Instantiated int/float case" << std::endl;
    }
};

template<> struct S<int>: S_int_or_float {};
template<> struct S<float>: S_int_or_float {};

Edit: , ... , ... , :

struct ParentS { /* some implementation */ }; 

template <class First, class Second>
struct S<First, Second, enable_if_t<trait_for_first_and_second<First, Second>::value, first_accepted_type_of_third_parameter> >: ParentS { };

template <class First, class Second>
struct S<First, Second, enable_if_t<trait_for_first_and_second<First, Second>::value, second_accepted_type_of_third_parameter> >: ParentS { };

// ...

, SFINAE, ...