Indirect forwarding references

Question

It is well known that \"direct\" forwarding references works in an easy way :

template
void f(T &&t); // Here we are.

Answer 1

No, this is not possible. If you want to constrain the function to only accept some_class_template, you have to use a type trait:

template <typename T>
struct is_some_class_template
    : std::false_type
{};

template <typename T>
struct is_some_class_template<some_class_template<T>>
    : std::true_type
{};

The trait can be used in several ways. One is with SFINAE:

template <typename T,
    std::enable_if_t<is_some_class_template<std::decay_t<T>>::value, int> = 0>
void f(T&& f); // f is some_class_template<U> for some U

Or you may find tag dispatch to be preferable:

template <typename T>
void f_impl(T&& f, std::true_type); // f is some_class_template<U> for some U

template <typename T>
void f_impl(T&& f, std::false_type); // f is something else

template <typename T>
void f(T&& f) {
    return f_impl(std::forward<T>(f), is_some_class_template<std::decay_t<T>>{});
}