Распаковка parameter pack в наследуемых типах

#include <iostream>
#include <utility>
 
struct A {int val; void foo() const {std::cout << val;}};
struct B {int val; void foo() const {std::cout << val;}};
struct C {int val; void foo() const {std::cout << val;}};
struct D {int val; void foo() const {std::cout << val;}};
 
template <typename ... Types>
struct Test : public Types ...
{
   Test() = default;
   Test(Types&& ...args) : Types(std::forward<Types>(args))... {}
 
   void print() const
   {
      Types::foo()...;
   }
 
};
 
int main()
{
   Test<A, B, C, D> t{{1}, {2}, {3}, {4}};
   t.print();
}

    void print() const {
        using dummy = int [];
        dummy{ (Types::foo(), 0)... };
    }

template <typename ... STR>
void foo(STR&& ... strs)
{
    using dummy = int[];
    dummy {(static_assert(std::is_convertible<STR, std::string>::value,
                  "arguments must be convertible to string"), 0)...} ;
}
 
int main()
{
    std::string ext = ".dat";
    foo(".txt", ext, ".bat");
}

template <typename ... STR>
void foo(STR&& ... strs)
{
    using com_t = std::common_type_t<STR...>;
    static_assert(std::is_convertible<com_t, std::string>::value,
                  "arguments must be convertible to string");
}