Universal reference to template template parameter

Question

Is it possible to pass a template template paramater value by universal reference? Consider for example this minimal example for a function (not) working on STL sequences:

#include <iostream>
#include <vector>

template < template<typename,typename> class C, template<typename> class A, typename T >
void func(C<T, A<T>>&& c) {
    // usually I'd std::forward here, but let's just use cout...
    std::cout << c.size() << "\n";
}

int main (int argc, char const* argv[]) {
    func(std::vector<float>(2));
    std::vector<float> lv(3);
    func(lv);
}

It won't compile since the compiler doesn't know how to bind the l-value ("lv") in the second call to func. I'm a bit lost when it comes to the deduction rules for the type of C. Can anyone enlighten me?

Edit: Although I guess it is not relevant for the question: I used g++ 4.9 and clang 3.5 (both repo HEADs)

Answer 1

"Universal reference" is a colloquialism, and it always means, strictly, a reference T && where T is a deduced template parameter.

You can modify your code, though, to use just that:

template <typename T>
void func(T && c) { /* ... std::forward<T>(c) ... */ }

If you care that T is of the specified form, add a trait:

#include <type_traits>

template <typename T>
typename std::enable_if<IsATemplate<typename std::decay<T>::type>::value>::type
func(T && c) { /* ... std::forward<T>(c) ... */ }

Writing the IsATemplate trait is left as an exercise.