Deducing knowledge of original types, whilst simultaneously forwarding

Summary: I want to end up with a function that deduces the exact types it was called with and takes (e.g.) a tuple that forwards them (the types of which will be different from the exact types the function was called with).

I'm stuck trying to "know" via deduction the types of the arguments to a given function, whilst simultaneously forwarding them. I think I might be missing something crucial about how this works.

#include <tuple>
#include <string>
#include <functional>

template <typename ...Args>
struct unresolved_linker_to_print_the_type {
   unresolved_linker_to_print_the_type();
};

void f(int,double,void*,std::string&,const char*) {
}

template <typename F, typename ...Args>
void g1(F func, Args&&... args) {
  unresolved_linker_to_print_the_type<Args...>();
  auto tuple = std::forward_as_tuple(args...);
  unresolved_linker_to_print_the_type<decltype(tuple)>();
}

template <typename F, typename T, typename ...Args>
void g2(F func, const T& tuple, Args... args) {
  unresolved_linker_to_print_the_type<Args...>();
  unresolved_linker_to_print_the_type<decltype(tuple)>();
}

int main() {
  int i;
  double d;
  void *ptr;
  std::string str;
  std::string& sref = str;
  const char *cstr = "HI";

  g1(f, i,d,ptr,sref,cstr);
  g2(f, std::forward_as_tuple(i,d,ptr,sref,cstr),  i,d,ptr,sref,cstr);
}

What I'd like to see is a scenario where when my function (e.g. g1 or g2) gets called it knows and can use both the original types - int,double,void*,std::string&,const char* and the forwarded arugments too.

In this instance I don't seem to be able to find this information from within g1 or g2. The (deliberate, to print out the types) linker error shows me in g1 they are:

int&, double&, void*&, std::string&, char co