Gcc is rightly warning you. Despite it's appearances (it takes Base argument), it is not a function template.
Your class definition has a non-template declaration of the friend function (without the template), but the friend function definition later on is a function template (i.e. starts with template..).
Also your operator<< takes a Base *. This is not correct. It should be Base const & to retain it's built-in semantics
Probably you are looking at something as below:
template <typename T>
class Base {
public:
friend ostream& operator << (ostream &out, Base<T> const &e){
return out;
};
};
int main(){
Base<int> b;
cout << b;
}
If you want fully templated, then this is probably what you want. But I am not sure how much useful this is over the previous one. Since the lookup involves ADL, you will never be able to resolve to any condition where T is not equal to U (as long as the call is from a context not related to this class e.g. from 'main' function)
template <typename T>
class Base {
public:
template<class U> friend ostream& operator << (ostream &out, Base<U> const &e){
return out;
};
};
int main(){
Base<int> b;
cout << b;
}
