[c++] Operator overloading on class templates

I'm having some problems defining some operator overloads for template classes. Let's take this hypothetical class for example.

template <class T>
class MyClass {
  // ...
};

• operator+=

  // In MyClass.h
  MyClass<T>& operator+=(const MyClass<T>& classObj);
  
  
  // In MyClass.cpp
  template <class T>
  MyClass<T>& MyClass<T>::operator+=(const MyClass<T>& classObj) {
    // ...
    return *this;
  }
  

  Results in this compiler error:

  no match for 'operator+=' in 'classObj2 += classObj1'
  

• operator<<

  // In MyClass.h
  friend std::ostream& operator<<(std::ostream& out, const MyClass<T>& classObj);
  
  
  // In MyClass.cpp
  template <class T>
  std::ostream& operator<<(std::ostream& out, const MyClass<T>& classObj) {
      // ...
      return out;
  }
  

  Results in this compiler warning:

  friend declaration 'std::ostream& operator<<(std::ostream&, const MyClass<T>&)' declares a non-template function
  

What am I doing wrong here?

This helped me with the exact same problem.

Solution:

1. Forward declare the friend function before the definition of the class itself. For example:

      template<typename T> class MyClass;  // pre-declare the template class itself
      template<typename T> std::ostream& operator<< (std::ostream& o, const MyClass <T>& x);
   

2. Declare your friend function in your class with "<>" appended to the function name.

      friend std::ostream& operator<< <> (std::ostream& o, const Foo<T>& x);
   

This way works:

class A
{
    struct Wrap
    {
        A& a;
        Wrap(A& aa) aa(a) {}
        operator int() { return a.value; }
        operator std::string() { stringstream ss; ss << a.value; return ss.str(); } 
    }
    Wrap operator*() { return Wrap(*this); }
};

You need to say the following (since you befriend a whole template instead of just a specialization of it, in which case you would just need to add a <> after the operator<<):

template<typename T>
friend std::ostream& operator<<(std::ostream& out, const MyClass<T>& classObj);

Actually, there is no need to declare it as a friend unless it accesses private or protected members. Since you just get a warning, it appears your declaration of friendship is not a good idea. If you just want to declare a single specialization of it as a friend, you can do that like shown below, with a forward declaration of the template before your class, so that operator<< is regognized as a template.

// before class definition ...
template <class T>
class MyClass;

// note that this "T" is unrelated to the T of MyClass !
template<typename T>
std::ostream& operator<<(std::ostream& out, const MyClass<T>& classObj);

// in class definition ...
friend std::ostream& operator<< <>(std::ostream& out, const MyClass<T>& classObj);

Both the above and this way declare specializations of it as friends, but the first declares all specializations as friends, while the second only declares the specialization of operator<< as a friend whose T is equal to the T of the class granting friendship.

And in the other case, your declaration looks OK, but note that you cannot += a MyClass<T> to a MyClass<U> when T and U are different type with that declaration (unless you have an implicit conversion between those types). You can make your += a member template

// In MyClass.h
template<typename U>
MyClass<T>& operator+=(const MyClass<U>& classObj);


// In MyClass.cpp
template <class T> template<typename U>
MyClass<T>& MyClass<T>::operator+=(const MyClass<U>& classObj) {
  // ...
  return *this;
}

You must specify that the friend is a template function:

MyClass<T>& operator+=<>(const MyClass<T>& classObj);

See this C++ FAQ Lite answer for details.