Access base class protected members using pointers to base class in derived class

Question

Consider the following code:

#include <iostream>

using std::endl;
using std::cout;

template<typename T>
class B{
protected:
    T value;
    B* ptr;
public:
    B(T t):value(t), ptr(0){}
};

template<typename T>
class D: public B<T>{
public:
    void f();
    D(T t):B<T>(t){}
};

template<typename T>
void D<T>::f(){
    cout << this->value << endl;     //OK!
    this->ptr = this;
    cout << this->ptr->value << endl;      //error! cannot access protected member!!
    B<T>* a = this;
    cout << a->value <<endl;       //error! cannot access protected member!!
}


int main(){
    D<double> a(1.2);
    a.f();
    return 0;
}

It seems that the member of the base class can be directly accessed using this pointer, but not the other pointers.
Does the compiler consider them as different instantiation?

Answer 1

Yes, this is expected behavior. Protected members of base class can be accessed in the derived class, but only via an object of a type that is the derived class (or the further derived class of current derived class) (including this). That means you can't access protected members via a pointer to base class.

A protected member of a class Base can only be accessed

1) ...

2) by the members of any class derived from Base, but only when operating on an object of a type that is derived from Base (including this)