C++: Why am I getting the error "expected type-specifier" when using declval as a template argument?

Here is the function I'm dealing with:

    template <typename T, typename C = std::declval<T>> T & f(C c = C()) {

        return c;

    }

It's supposed to run std::declval[1] through the function f. I've pinpointed the problem down to the the default template argument. It seems the compiler is not interpreting std::declval as a type. Why? I'm getting the following errors:

    prog.cpp:17:36: error: expected type-specifier

    prog.cpp:17:36: error: expected '>'

    prog.cpp: In function 'int main()':

    prog.cpp:23:10: error: no matching function for call to 'f()'

    prog.cpp:23:19: error: no matching function for call to 'f()'

So I tried using decltype on it:

    template <typename T, typename C = decltype( std::declval<T> )> T & f(C c = C()) {

        return c;

    }

The errors go away until I have to call f:

    f<A>();

(Where A is an arbitrary class; remember that I'm trying to run std::declval through the function; so try and think of the function f as std::declval<A>).

But I get a large amount of template errors, most of them saying that C is A && when I call f with f<A>(). I never wanted C to be an rvalue reference, so then I try to take out the reference using std::remove_reference[2]:

    template <typename T, typename C = std::remove_reference<decltype( std::declval<T> )>::type> T & f(C c = C()) {

        return c;

    }

But then I get the exact same errors. What could I be doing wrong? Why isn't std::declval<T> deduced as a type? If you guys need context, here's what I have:

    #include <utility>

    #include <type_traits>

    struct B;

    struct A {

        void operator *(const B &) {}

    };

    struct B {

        void operator *(int x) const {}

    };

    inline void operator *(int a, const B & b) {

        return b * a;

    }

    template <typename T, typename C = std::remove_reference<decltype( std::declval<T> )>::type> T & f(C c = C()) {

        return c;

    }

    int main() {

        f<A>() * f<B>();

    }

The above code surrounding the function does nothing special. What you should be focusing on is why std::declval<T> is not a type and how I can fix this. Here is a demo:

    http://ideone.com/Zluu3d#view_edit_box

All answers and suggestions are appreciated.

    [1]: Found in header <utility>. Converts any type T into a reference type, thus allowing the ability to use member functions without specifying constructors

    [2]: Found in header <type_traits>. Removes a given reference