declare template friend function of template class

Ryan Haining picture Ryan Haining · Sep 13, 2013 · Viewed 27.5k times · Source

I have a class template Obj and a function template make_obj. Obj has a private single constructor defined, which takes a reference to its templated type to bind to.

template <typename T>
class Obj {
  private:
    T& t;
    Obj(T& t)
        : t{t}
    { }
};

template <typename T>
Obj<T> make_obj(T& t) { 
    return {t};
}

What I want is to declare the make_obj function a friend so that it can create Obj's, but no one else can (except via the copy ctor).


I have tried several friend declaration including

friend Obj make_obj(T&);

and

template <typename T1, typename T2>
friend Obj<T1> make_obj(T2&);

The latter being a less than desirable attempt at making all template instantiations of make_obj friends of the Obj class. However in both of these cases I get the same error:

error: calling a private constructor of class 'Obj<char const[6]>'
    return {t};
           ^

note: in instantiation of function template specialization
      'make_obj<const char *>' requested here
    auto s = make_obj("hello");
             ^

trying to do make_obj("hello"); for example purposes.

How can I allow only make_obj access to Obj's value contructor?

Answer

Daniel Frey picture Daniel Frey · Sep 13, 2013

You need a few forward declarations:

template <typename T>
class Obj;

template <typename T>
Obj<T> make_obj(T t);

template <typename T>
class Obj {
private:
    T & t;
    Obj (T & t) : t(t) { }
    Obj() = delete;

    friend Obj make_obj<T>(T t);
};

template <typename T>
Obj<T> make_obj(T t) { 
    return Obj<T>(t);
}

live example

And BTW: I don't think you really want T & t; for your class' member variable. Probably T t; is a better choice ;)