Universal reference to template template parameter

c++, c++11, templates

Solution

"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.

Problem

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)

Original source