C++ Access derived class member from base class pointer

Question

class Base
{
    public:
    int base_int;
};

class Derived : public Base
{
    public:
    int derived_int
};

Base* basepointer = new Derived();
basepointer-> //Access derived_int here, is it possible? If so, then how?

Answer 1

No, you cannot access derived_int because derived_int is part of Derived, while basepointer is a pointer to Base.

You can do it the other way round though:

Derived* derivedpointer = new Derived;
derivedpointer->base_int; // You can access this just fine

Derived classes inherit the members of the base class, not the other way around.

However, if your basepointer was pointing to an instance of Derived then you could access it through a cast:

Base* basepointer = new Derived;
static_cast<Derived*>(basepointer)->derived_int; // Can now access, because we have a derived pointer

Note that you'll need to change your inheritance to public first:

class Derived : public Base

Answer 2

You're dancing on the minefield here. The base class can never know that it's actually an instance of the derived. The safest way to do that would be to introduce a virtual function in the base:

class Base 
{ 
protected:
 virtual int &GetInt()
 {
  //Die horribly
 }

public: 
 int base_int; 
}; 

class Derived : Base 
{ 
  int &GetInt()
  {
    return derived_int;
  }
public: 
int derived_int 
}; 

basepointer->GetInt() = 0;

If basepointer points as something other that a Derived, your program will die horribly, which is the intended result.

Alternatively, you can use dynamic_cast(basepointer). But you need at least one virtual function in the Base for that, and be prepared to encounter a zero.

The static_cast<>, like some suggest, is a sure way to shoot yourself in the foot. Don't contribute to the vast cache of "unsafety of the C language family" horror stories.