Transparent constexpr int as type identifier

I have a struct Foo

. I would like to add some kind of identifier (placeholder?) To statically select from a tuple of values ​​passed to various Foo

objects. Ideally, I would not become a template type because it causes a lot of changes elsewhere.

I tried adding an integer to it and using constexpr expressions (Demo)

#include <tuple>

using namespace std;

struct Foo {
private:
  const int pl;

public:
  constexpr Foo (int c) : pl(c) {}

  constexpr int place() {return pl;}

  template <typename... T>
  constexpr int extract(std::tuple<T ...> const &vals) {
    // I would like to uncomment the following but it fails compiling
    // return std::get<pl>(vals);
    return std::get<0>(vals);
  }
};

int main(void) {
   constexpr Foo f(1);
   constexpr std::tuple<int, int> test = std::make_tuple(0, 10);

   // The following passes
   static_assert(f.place() == 1, "ERROR");
   // The following fails
   static_assert(f.extract(test) == 10, "ERROR");
}

      

        
        
        
      

    

I was expecting to be able to use constexpr place()

in get<>

, what am I missing?

Do I have a way out without using templates?