Question

In code:

template<class T>
struct is_builtin
{
    enum {value = 0};
};

template<>
struct is_builtin<char>
{
    enum {value = 1};
};

template<>
struct is_builtin<int>
{
    enum {value = 1};
};

template<>
struct is_builtin<double>
{
    enum {value = 1};
};

template<class T>
struct My
{
    typename enable_if<is_builtin<T>::value,void>::type f(T arg)
    {
        std::cout << "Built-in as a param.\n";
    }


    typename enable_if<!is_builtin<T>::value,void>::type f(T arg)
    {
        std::cout << "Non - built-in as a param.\n";
    }
};

struct A
{
};

int main()
{
    A a;
    My<int> m;
    My<A> ma;
    m.f(1);
    ma.f(a);
    return 0;
}

I’m getting an error:

error C2039: 'type' : is not a member of 'std::tr1::enable_if<_Test,_Type>'    

Obviously I don’t understand how to use enable_if. What I was thinking was that I can enable one or the second one member function from a set of member functions during compilation time but it does not work. Could anyone please explain to me how to do it correctly?
Edited
What I really can’t understand is why isn’t there typedef in one of those def. Compiler cannot find it and it wont compile it.

Answer 1

You can’t use class template parameters to get SFINAE for member functions.

You either need to

• make the member function a member function template instead and use enable_if on the member function template’s template parameters or

• move the member function f into a policy class and specialize the class template using enable_if.