C++ pointer multi-inheritance fun.

Question

I'm writing some code involving inheritance from a basic ref-counting pointer class; and some intricacies of C++ popped up. I've reduced it as follows:

Suppose I have:

class A{};
class B{};
class C: public A, public B {};

C c;
C* pc = &c;
B* pb = &c;
A* pa = &c;

// does pa point to a valid A object?
// does pb point to a valid B object?

// does pa == pb ?

Furthermore, does:

// pc == (C*) pa ?
// pc == (C*) pb ?

Thanks!

Answer 1

• does pa point to a valid A object?
• does pb point to a valid B object?

Yes, the C* gets converted so that pa and pb point to the correct addresses.

• does pa == pb ?

No, usually not. There can't be an A object and a B object at the same address.

Furthermore, does

• pc == (C*) pa ?
• pc == (C*) pb ?

The cast converts the pointers back to the address of the C object, so both equalities are true.

Answer 2

pc == pa;
pc == pb;

Not defined, depends on class structure.

pc == (C*) pa;
pc == (C*) pb;

Thats ok.

pa == pb;

No.

Do they point to valid objects?

Yes

Answer 3

C embeds an A and a B.

class C: public A, public B {};

is very similar to the C code

struct C {
    A self_a;
    B self_b;
};

and (B*) &c; is equivalent to static_cast< B* >( &c ) is similar to &c.self_b if you were using straight C.

In general, you can't rely on pointers to different types being interchangeable or comparable.

Answer 4

What you get is something like this in memory

 ----------
 | A data |
 ----------
 | B data |
 ----------
 | C data |
 ----------

So if you want the entire C object you'll get a pointer to the beginning of the memory. If you want only the A "part", you get the same address since that's where the data members are located. If you want the B "part" you get the beginning + sizeof(A) + sizeof(whatever the compiler adds for vtable). Thus, in the example, pc != pb (could be pc != pa) but pa is never equal to pb.