C++ template inheritance hides template parameters

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Category:Languages
template <typename> struct B {     constexpr static int T = 5; };  template <int T> struct D : B<int> {     constexpr static int value = T; };  int main() {     std::cout << D<7>::value << std::endl; } 

As I have recently learned that template parameters of a template derived class gets checked after the names in the base class scope during lookup. That being said, is there anyway to qualify value to refer to the template parameter T in the derived class?

 


I'm not entirely sure I understand the question but I think decltype does what you want:

template <int T> struct D : B<decltype(T)> {     constexpr static decltype(T) value = T; }; 

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