Deducing template member function in class template

Question

Why can't T be deduced from fn's signature in the following example?

template<int I>
struct Class {
    template<typename T>
    Class &operator <<( void (*)(const Class<I> &, const T &) ) { return *this; }
};

struct Ot { };

template<int I>
void fn(const Class<I> &, const Ot &) { }

int main() {
    Class<1>() << fn;
}

Where in contrast, the following example without the operator<< being a regular member is legal:

template<int I>
struct Class {
    Class &operator <<( void (*)(const Class<I> &) ) { return *this; }
};

struct Ot { };

template<int I>
void fn(const Class<I> &) { }

int main() {
    Class<1>() << fn;
}

Answer 1

Because just like you're using Class<1> instead of Class, you have to provide the template parameter to the function template fn as well:

Class<1>() << fn<1>;

Also, you might want to return a reference from your operator <<:

template<typename T>
Class & operator <<( void (*)(const Class<I> &, const T &) ) { return *this; }
      ^