Templates and overloadering operator << [duplicate]

Answer 1

friend std::ostream& operator<<(std::ostream& os, const Derived& dt);

declares a non template version friend, so you would have to implement

std::ostream& operator<<(std::ostream& os, const Derived& dt);

for every T you use:

std::ostream& operator<<(std::ostream& os, const Derived& dt) {
    return os << dt.data_;
}

And so on. A way to do it without duplicate code is with definition inside the class:

template 
class Derived : public IBase
{
public:
    explicit Derived(T data);
    friend std::ostream& operator<<(std::ostream& os, const Derived& dt)
    {
        return os << dt.data_;
    }
private:
    T data_;
};

Demo

An other alternative is to make the function template.

// Forward declarations
template  class Derived;
template  std::ostream& operator<<(std::ostream& os, const Derived& dt);

And then in Derived, you have 2 choices:

• Declare each the template method friend:

  template 
  friend std::ostream& operator<<(std::ostream& os, const Derived& dt);
  

  So std::ostream& operator<<(std::ostream& os, const Derived& dt) has access to private members of Derived, but also to the ones of Derived

  Demo

• Declare only the template which matches the argument:

  friend std::ostream& operator<< <>(std::ostream& os, const Derived& dt);
  

  and so std::ostream& operator<<(std::ostream& os, const Derived& dt) doesn't have access to private members of Derived.

  Demo