Giving Variadic Template Pack as function specification

I'm trying to create a function that has a template pack specialization (it has no parameters) and it prints a message, until the last one where function specialization makes it print something else, and stop. As I'm really bad explaining , here I post the code of what I'm trying to do:

template <typename T>
constexpr void UpdateStuff()
{
    std::cerr << "I am the last one :D" << std::endl;

}
template< typename T ,typename... TT>
constexpr void UpdateStuff()
{



    std::cerr << "I am NOT the last one :D";

    UpdateStuff<TT...>();

}
int main()
{


    UpdateStuff<int,double>(); // Should only print text twice
    std::cin.get();

    return 1;
}

As a first note , I know this does not work, UpdateStuff<TT...>(); it yields an ambigous call to overloaded function , I've managed to get it working by giving the function UpdateStuff() parameters like UpdateStuff(T first, TT... second) and the specialization only one UpdateStuff(T first) but I want to know if this is possible without function parameters , I'll give a summary of my questions:

• I dont completely understand why UpdateStuff<TT...>(); doesn't work, if TT is just double at compile-time why doesnt it use the first function?
• Is it possible to do what I am trying to accomplish without function parameters?

Thanks in advance , and if you don't understand something related to my question I'd be grateful to explain it better,and sorry for my poor english.

Yes, it's possible.

But take in count that typename ... TT is "zero or more typenames", so calling UpdateStaff<double>() both version of UpdateStaff() are ok.

You can, by example, impose a second type in the not final version; something like this

template <typename T1, typename T2, typename... TT>
constexpr void UpdateStuff()
{
    std::cerr << "I am NOT the last one :D";

    UpdateStuff<T2, TT...>();
}