Can a friend class object access base class private members on a derived class object?

Question

I\'m surprised that the code below compiles.

It seems that a class befriended to the (publicly inherited) base class can access a member of the base class provided an in

Answer 1

1. It seems that somehow friendship is inherited and a friend class can access a member of the derived class.
   In short, how is d.b_var valid within F::func(D& d)?

d.b_var might be misleading. To be more precise (another way to see it), b_var is not (direct) member of derived class D. Instead, the object of D contains a subobject of base class B, which has member b_var, and could be accessed by the friend F. (As we can also write d.b_var as d.B::b_var.)

$10/3 Derived classes [class.derived]:

The base-specifier-list specifies the type of the base class subobjects contained in an object of the derived class type. [ Example:

struct Base {
  int a, b, c;
};

struct Derived : Base {
  int b;
};

struct Derived2 : Derived {
  int c;
};

Here, an object of class Derived2 will have a subobject of class Derived which in turn will have a subobject of class Base. — end example ]

And

2. If the inheritance is changed to private then compilation fails.

Because

class B {
    int b_var;
    friend class F;
};

class D: private B {
    int d_var;
};

class F{
public:
    void func(D &d) {
        d.b_var = 5;  // Fail. Can't access subobject of B
        d.d_var = 5;  // Fail. Can't access member of D
    }
};

Then

class B {
    int b_var;
};

class D: private B {
    friend class F;
    int d_var;
};

class F{
public:
    void func(D &d) {
        d.b_var = 5;  // Fail. Can't access b_var of subobject of B
        d.d_var = 5;  // Fine.
    }
};

Note that in last case, even F is friend of class D, it could access all the private/protected members of D, but not including members in subobject B, because they're not (direct) members of class D.