Variadic template function with no arguments

Question

I want to write a function that will do an operation based on the types, not arguments, of a function. As such, the function receives no template based arguments. The general gist is something like this:

#include <iostream>

void func() {
    std::cout<<"End marker\n";
}

template <typename Type, typename... T>
void func() {
    std::cout<<"Type sizeof "<<sizeof(T)<<"\n";

    func<T...>();
}

int main() {
    func<int, int, int>();
}

Which, of course, doesn't compile. I've tried doing this:

template <typename Type, typename... T>
void func() {
    std::cout<<"Type sizeof "<<sizeof(T)<<"\n";

    if( sizeof...(T)!=0 )
        func<T...>();
}

This does not work, however. The func<T...> may not get evaluated, but it does need to be compilable.

Is there a way to do this that I'm missing?

Answer 1

You can make your original setup work by making func a "template function" (that doesn't actually use the template) like:

template<int = 0>
void func() {
    std::cout<<"End marker\n";
}

template <typename Type, typename... T>
void func() {
    std::cout<<"Type sizeof "<<sizeof(Type)<<"\n";

    func<T...>();
}

And your second one can work by using if constexpr, so func<>() isn't compiled.