(Partially) specializing a non-type template parameter of dependent type

Question

Maybe I'm tired, but I'm stuck with this simple partial specialization, which doesn't work because non-type template argument specializes a template parameter with dependent type 'T':

template <typename T, T N> struct X;
template <typename T>      struct X <T, 0>;

Replacing 0 by T(0), T{0} or (T)0 doesn't help. So is this specialization even possible?

Answer 1

See paragraph [temp.class.spec] 14.5.5/8 of the standard:

The type of a template parameter corresponding to a specialized non-type argument shall not be dependent on a parameter of the specialization. [ Example:

template <class T, T t> struct C {};
template <class T> struct C<T, 1>; // error

template< int X, int (*array_ptr)[X] > class A {};
int array[5];
template< int X > class A<X,&array> { }; // error

—end example ]

The answer to your edit: the easiest workaround is to replace a non-type template parameter with a type one:

#include <type_traits>

template <typename T, typename U>
struct X_;

template <typename T, T N>
struct X_<T, std::integral_constant<T, N>> {};

template <typename T>
struct X_<T, std::integral_constant<T, 0>> {};

template <typename T, T N>
struct X : X_<T, std::integral_constant<T, N>> {};

Answer 2

Solution using Yakk's solution:

#include <iostream>
#include <type_traits>

template <typename T, T N, typename = void > 
struct X {
  static const bool isZero = false;
};

template <typename T, T N>
struct X < T, N, typename std::enable_if<N == 0>::type > {
  static const bool isZero = true;
};

int main(int argc, char* argv[]) {
    std::cout << X <int, 0>::isZero << std::endl;
    std::cout << X <int, 1>::isZero << std::endl;
    return 0;
}

Live Demo

Answer 3

You can add a typename=void parameter to the end of the list of template arguments, then go hog wild with std::enable_if_t< condition > in specializations.

Answer 4

You need to pass an integral value in a template, Both, your first and second template, will not work if the type T is not an integral type.

You can pass Traits as a typed template parameter to specify the value N:

#include <iostream>

// error: ‘double’ is not a valid type for a template non-type parameter
template <typename T, T N> struct X0;

// error: ‘double’ is not a valid type for a template non-type parameter
template <typename T, T N, int = 0> struct X1;



template <typename T, T N>
struct IntegralTraits {
    static constexpr T Value() { return N; }
};

template <typename T, typename Traits = void>
struct X2 {
    static constexpr T Value() { return Traits::Value(); }
};

template <typename T>
struct X2<T, void> {
    static constexpr T Value() { return T(); }
};


int main() {
    // error: ‘double’ is not a valid type for a template non-type parameter
    // X0<double, 0>();

    // error: ‘double’ is not a valid type for a template non-type parameter
    // X1<double, 0>();

    X2<int> a;
    X2<double, IntegralTraits<int, 1>> b;

    std::cout.precision(2);
    std::cout << std::fixed  <<  a.Value() << ", "<< b.Value() << '\n';
    return 0;
}

If you limit yourself to integral types pick a large one:

template <typename T, std::size_t N = 0> struct X {};