Is it possible to access values of non-type template parameters in specialized template class?

Question

Is it possible to access values of non-type template parameters in specialized template class?

If I have template class with specialization:

   template <int major, int minor> struct A {
       void f() { cout << major << endl; }
   }

   template <> struct A<4,0> {
       void f() { cout << ??? << endl; }
   }

I know it the above case it is simple to hardcode values 4 and 0 instead of using variables but what I have a larger class that I'm specializing and I would like to be able to access the values.

Is it possible in A<4,0> to access major and minor values (4 and 0)? Or do I have to assign them on template instantiation as constants:

   template <> struct A<4,0> {
       static const int major = 4;
       static const int minor = 0;
       ...
   }

Answer 1

This kind of problem can be solved by having a separate set of "Traits" structs.

// A default Traits class has no information
template<class T> struct Traits
{
};

// A convenient way to get the Traits of the type of a given value without
// having to explicitly write out the type
template<typename T> Traits<T> GetTraits(const T&)
{
    return Traits<T>();
}

template <int major, int minor> struct A 
{ 
    void f() 
    { 
        cout << major << endl; 
    }   
};

// Specialisation of the traits for any A<int, int>
template<int N1, int N2> struct Traits<A<N1, N2> >
{
    enum { major = N1, minor = N2 };
};

template <> struct A<4,0> 
{       
    void f() 
    { 
        cout << GetTraits(*this).major << endl; 
    }   
};

Answer 2

Not really an answer to your question, but you could enumerate them, viz:

enum{
 specialisationMajor=4,
 specialisationMinor=0
};

template <> struct A<specialisationMajor,specialisationMinor> {
    static const int major = specialisationMajor;
    static const int minor = specialisationMinor;
    ...
}

Answer 3

Not really an answer to your question, but the idea below helped me once:

#include <iostream>

template <int major, int minor, int= major, int= minor> struct A {
    void f() { std::cout << major << '\n'; }
};

template <int major, int minor> struct A<major, minor, 4, 0> {
    void f() { std::cout << major << ':' << minor << '\n'; }
};

int main()
{
    A<3, 3>().f();
    A<4, 0>().f();
}

Answer 4

No, you don't have access to the specialized non-type template parameters. But here is a way to not repeat yourself in the implementation of f:

   template <int major, int minor>
   struct f_impl
   {
       void f() { cout << major << endl; }
   };

   template <int major, int minor>
   struct A : public f_impl<major, minor>
   {};

   template <> struct A<4,0> : public f_impl<4,0>
   {};

For this example, one doesn't gain all too much, as one needs to write the 4,0 twice (--so you could write it as well a second time in the cout in your OP). But it starts to pay out if you have more functions using the template parameters.