I can't store the address of Derived class in the pointer of base class when inheritance is private, but when I inherit it in public it shows no error

Question

I am stuck with this code, when I store the address of the the Derived class in Pointer of base class, it shows error, but when made inheritance public there is no error, can anyone help..?

#include <iostream>
using namespace std;
class Base // Created a Class Base
{
public: 
void show()
{
cout << "base";
}
};
class Derived: private Base
{
public:
int d;
void display()
{
    cout << "derived";
}
};
int main()
{
Base b, *bptr;
Derived d, *dptr;
bptr = &b;
dptr = &d;
bptr->show();
bptr = &d;
bptr->show();
return 0;
}

@Nikhil - Because it is `private`. Private inheritance means that the class itself knows about its base, but nobody else is allowed to see that - that information is private — BoP, Dec 06 '21 at 09:07

Jason · Accepted Answer · 2021-12-06T04:33:10.660

In your case the class Base is not accessible and

$11.2/5 states -

If a base class is accessible, one can implicitly convert a pointer to a derived class to a pointer to that base class (4.10, 4.11). [ Note: it follows that members and friends of a class X can implicitly convert an X* to a pointer to a private or protected immediate base class of X. —end note ]

Since Base is not an accessible class of Derived when accessed in main, the Standard conversion from Derived class to Base class is ill-formed. Hence the error.

I can't store the address of Derived class in the pointer of base class when inheritance is private, but when I inherit it in public it shows no error

1 Answers1