How to write the definition of a derived class in c++?

Question

I don't know the syntax for writing the definition of a class function with templates.

I get an error, the complier expected ; , so that be void plus; Anyone knows to fix that!

template<class Type 
class basic{
 protected:
    Type x;
 public:
    int value;
    //virtual void  print() =0; 
  basic(Type xArg):x(xArg){}
}; 

template<class Type>
class plus:public basic<Type>{
 public: 
  Type y;
  plus(Type xArg, Type yArg):basic<Type>(xArg),y(yArg){}
  void print(); 

};

//template<class Type> 
void plus<Type>::print(){ 
  cout << "x : " << x << endl;
  cout << "y : " << y << endl;

}

Answer 1

Let's say we have a base class:

template<class T> 
class MyBase {
   public:
    T value;

    virtual void print() {
        std::cout << "MyBase.value: " << value;
    }        
    virtual ~MyBase() = default; 
};

We can derive it and override print like this. It doesn't have to be a template, but I'm using one in the example.

template<class T>
class MyDerived : MyBase<T>
{
   public:
    void print() override {
        std::cout << "MyDerived.value: " << MyBase<T>::value;
    }
};

In addition, something I like to do is use base to represent the base class:

template<class T>
class MyDerived : MyBase<T>
{
   public:
    using base = MyBase<T>; // Use this to indicate the base
    void print() override {
        std::cout << "MyDerived.value: " << base::value << '\n';
    }
};