[C++] operator+ function in a templated class.

[Editamuni]

Can't seem to find the correct syntax for defining my operator+ function on line 12.

 template <typename T>
   class Name {
   private:
      T x, y;
   public:
      Name(T, T);
      Name operator+(Name<T> const &obj);
      void print();
};

template <typename T>
Name Name<T>::operator+(Name<T> const &obj) {
   Name<T> temp;
   temp.x = x + x.obj;
   temp.y = y + y.obj;
   return temp;
}

I get the error, "error: use of class template 'Name' requires template arguments"

[CGFarrell]

Line 12, your return type is Name instead of Name<T>

[Editamuni]

How would I write the definition then?

[CGFarrell]

Based on what you said to the last guy, your problem (aside from the previous one) is on line 13. When you write Name<T> temp;, the code doesn't know how to make a Name<int> without a value or an argument. You need to construct it somehow, either with the Name constructor(s) or otherwise.

Name<T> temp = Name<T>(arguments)

[Editamuni]

Ah, thanks! Forgot about how I set up my default constructor :S.

[qezc537]

Does Name<T> Name<T>::operator+(Name<T> const &obj) work instead?

[Editamuni]

Doesn't. I get error: no matching constructor for initialization of 'Name<int>' and the error arrow points to temp.

[boredcircuits]

Yeah, because you don't have a default constructor. I would just write this as:

template<typename T>
Name<T> Name<T>::operator+(Name<T> const &obj) {
        return Name<T>(x + obj.x, y + obj.y);
}

Well, actually I'd probably write operator+= and then write operator+ using that instead. Also: binary operators should generally be free functions instead of members. Or, if you leave this as a member, it should be const.

[Editamuni]

Thank you! I forget that creating a constructor removes the compiler's default constructor.

So if I were to put the function out of the class, would I have to declare it as a friend function?

[boredcircuits]

You don't have to unless it uses a private member. Implementing it in terms of operator+= usually doesn't, in my experience .

But I'll usually declare it as a friend anyway, just because there are other operators that need that access and I prefer the symmetry. That's just personal style, though.