(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