template argument deduction with strongly-typed enumerations

Question

If I have a normal (weak) enumeration, I can use its enumerated values as non-type template parameters, like so:

enum { Cat, Dog, Horse };  template <int Val, typename T> bool magic(T &t) {     return magical_traits<Val>::invoke(t); } 

and call it as: magic<Cat>(t)

as far as I can see, if I have a strongly-typed enumeration and don't want to hard-code the enumeration type, I end up with:

enum class Animal { Cat, Dog, Horse };  template <typename EnumClass, EnumClass EnumVal, typename T> bool magic(T &t) {     return magical_traits<EnumVal>::invoke(t); } 

and now I have to write: magic<Animal, Animal::Cat>(t), which seems redundant.

Is there any way to avoid typing out both the enum class and the value, short of

#define MAGIC(E, T) (magic<decltype(E), E>(T)); 

Answer 1

You can do it like this, if you can use C++17

#include <type_traits>  enum class Animal { Cat, Dog, Horse };  template <typename EnumClass, EnumClass EnumVal>  void magic_impl() {     static_assert(std::is_same_v<EnumClass, Animal>);     static_assert(EnumVal == Animal::Cat); }  template <auto EnumVal> void magic() {     magic_impl<decltype(EnumVal), EnumVal>(); }  int main() {     magic<Animal::Cat>(); } 

demo: http://coliru.stacked-crooked.com/a/9ac5095e8434c9da

Answer 2

I'm sorry, I have to tell you that

It is not possible

Take the macro, put it into a scary named header and protect it from your colleague's cleanup script. Hope for the best.