Why compile error with enable_if

Question

Why this does not compile with gcc48 and clang32?

#include 

template  
struct S {

    template 
    typename         


        

Answer 1

Well, I am not sure what you wanted to do, but maybe this code will help:

#include <iostream>

template <int N>
struct S {

    template<class T=int>
    typename std::enable_if<N==1, T>::type
    f(T t) {return 1;}

    template<class T=int>
    typename std::enable_if<N!=1, T>::type
    f(T t) {return 2;}
};

int main()
{
    S<1> s1;
    S<2> s2;
    std::cout << s1.f(99) << " " << std::endl << s2.f(5);
}

This prints 1 and 2.

Answer 2

To get std::enable_if to work like this, you are relying on SFINAE. Unfortunately, at the point where you declare

S<1> s1;

it will instantiate all of S<1>'s member declarations. SFINAE will only come into play at this point if S<1> were an ill-formed construct. It is not. Unfortunately, it contains a function which is invalid, thus the instantiation of S<> is invalid.

For things like this, I might defer to a seperate template struct:

template <bool B>
struct f_functor {
    template <typename T>
    static int f(T t) { return 1; }
};

template <>
struct f_functor<false> {
    template <typename T>
    static int f(T t) { return 2; }
};

template <int N> 
struct S {

    template<class T> 
    typename int f(T t) { return f_functor<N==1>::f(t); }
};

Answer 3

For this case you could think about not using enable_if at all. It is posible to simply specialise f:

template <int N> 
struct S {
    template<class T> int f(T t);
};

template<int N>
template<class T>
int S<N>::f(T t) { return 2; }

template<>
template<class T>
int S<1>::f(T t) { return 1; }

int main() {
    S<1> s1;
    return s1.f(99);
}

Answer 4

Use a default boolean template parameter, like this:

template <int N> 
struct S {

    template<class T, bool EnableBool=true> 
    typename std::enable_if<N==1 && EnableBool, int>::type
    f(T t) {return 1;};

    template<class T, bool EnableBool=true> 
    typename std::enable_if<N!=1 && EnableBool, int>::type
    f(T t) {return 2;};
};

Answer 5

Because you use enable_if without using the template parameter T in your function templates. If you want to specialize for when the struct S has a certain template parameter value N, you'll need to use class template specialization.

template <int N, class Enable = void> 
struct S {  };

template <int N>
struct S<N, typename std::enable_if<N == 1>::type>
{
  ....
};