Question about Virtual Inheritance hierarchy - c++

Question

I encounter this problem when tackling with virtual inheritance. I remember that in a non-virtual inheritance hierarchy, object of sub-class hold an object of its direct super-class. What about virtual inheritance? In this situation, does object of sub-class hold an object of its super-class directly or just hold a pointer pointing to an object of its super-class?
By the way, why the output of the following code is:
sizeof(A): 8
sizeof(B): 20
sizeof(C): 20
sizeof(D): 36
Code:
#include <iostream>
using namespace std;
class A{
char k[ 3 ];
public:
virtual void a(){};
};
class B : public virtual A{
char j[ 3 ];
public:
virtual void b(){};
};
class C : public virtual A{
char i[ 3 ];
public:
virtual void c(){};
};
class D : public B, public C{
char h[ 3 ];
public:
virtual void d(){};
};
int main( int argc, char *argv[] ){
cout << "sizeof(A): " << sizeof( A ) << endl;
cout << "sizeof(B): " << sizeof( B ) << endl;
cout << "sizeof(C): " << sizeof( C ) << endl;
cout << "sizeof(D): " << sizeof( D ) << endl;
return 0;
}
Thanks in advance.
Kind regards.

Answer 1

The virtual base object is somewhere in the memory block that belongs to the object (the memory with size = sizeof(object)). Because several sub objects of different types can be combined in various ways but must share the same base object, a offset pointer is needed for each sub object to find out the virtual base object. Without virtual inheritance, the offset to find out the corresponding base object is fixed at compile time for each class type.
The sizeof values depend on your compiler and machine, but the following assumptions are very common:
assumption: pointer size is 4 bytes
assumption: class size is rounded up to multiple of 4 bytes
sizeof(A): 8 -> 1 pointer to vtable (virtual method)
+ 3 chars -> 4+3=7
-> round up to 8
sizeof(B): 20 -> 8 + 1 pointer to vtable (virtual method)
+ 1 offset pointer to virtual base
+ 3 chars -> 8 + 4 + 4 + 3 = 19
-> round up to 20
sizeof(C): 32 -> 20 + 1 pointer to vtable (virtual method)
+ 1 offset pointer to virtual base
+ 3 chars
-> 20 + 4 + 4 + 3 = 31 // this calculation refers to an older
-> round up to 32 // version of the question's example
// where C had B as base class
The calculations are guessed because the real calculation must exactly know how the compiler works.
Regards,
Oliver
More details why an extra offset pointer is needed:
Example:
class B : virtual public A {...};
class C : virtual public A {...};
class D1 : public B {...};
class D2 : public B, C {...};
possible memory layout for D1:
A
B
D1
possible memory layout for D2:
A
C
B
D2
in the second case sub object B needs another offset to find its base A
An object of type D2 consists of a memory block, where all the parent object parts are contained, i.e. the memory block for an object of type D2 has a section for the A member variables, the C member variables, the B member variables and the D2 member variables. The order of these sections is compiler dependent, but the example shows, that for multiple virtual inheritance a offset pointer is needed, that points within the object's total memory block to the virtual base object. This is needed because the methods of class B know only one this pointer to B and must somehow calculate where the A memory part is relative to the this pointer.
Calculation sizeof(D):
sizeof(D): 36 -> A:3 chars + A:vtable
+ B:3 chars + B:vtable + B:virtual base pointer
+ C:3 chars + C:vtable + C:virtual base pointer
+ D:3 chars + D:vtable
= 3 + 4
+ 3 + 4 + 4
+ 3 + 4 + 4
+ 3 + 4
= 36
The above calculation is probably wrong ;-) ...
I'm not sure whether the D part has its own vtable pointer or not (this is all highly compiler dependent).
I now think that it could be that the D part use the vtable pointer entry of its parent classes and that the 4 extra bytes are used for alignment each part (multiple of 8 bytes):
So this calculation is probably more correct:
sizeof(D): 36 -> A:3 chars + A:vtable + A:alignment
+ B:3 chars + B:vtable + B:virtual base pointer + B:alignment
+ C:3 chars + C:vtable + C:virtual base pointer + C:alignment
+ D:3 chars + D:alignment
= 3 + 4 + 1
+ 3 + 4 + 4 + 1
+ 3 + 4 + 4 + 1
+ 3 + 1
= 36

Answer 2

I see three point analysis for the above question
a. Virtual Inheritance
"Virtual inheritance is a mechanism whereby a class specifies that it is willing to share the state of its virtual base class. Under virtual inheritance, only one, shared base-class subobject is inherited for a given virtual base regardless of how many times the class occurs as a virtual base within the derivation hierarchy. The shared base-class subobject is called a virtual base class." ... From Lippman
Virtual inheritance only avoids duplicate sub-objects inherited from multiple inheritance. But this does not indicate in any way that the base class objects will not be sub-objects. On the contrary, the sub-object (atleast one copy would be present - I mean would be included in sizeof() operation) even during the virtual inheritance.
b. virtual function
Virtual function is for dynamic binding of member functions of objects involved in hierarchy. So even this does not have any significance towards sub-object arrangements.
c. Implementation of the sub-objects
This is totally compiler dependent, and for all reasons would be very difficult to determine - in its implementation. However, we can confirm that the sizeof() of the object would include the size of the base class (sub) objects also - and we can visualize them as having the base class object embedded in them.
Each object of the inherited function will definitely contain space for the sub-objects.
HTH

Answer 3

does object of sub-class hold an object of its super-class directly
Yes, that is how it works whether the inheritance is virtual or not. I would use the word "contain" vs. "hold" however.
If your hierarchy looked like this, with no virtual inheritances anywhere:
# W <--- base class
# / \
# X Y <--- subclasses of W
# \ /
# Z <--- most derived class
Then Z will have two copies of W. But if you make the X-->W and Y-->W inheritances virtual, then Z will only have one copy of W because Z's two superclasses share their common base class.
# W
# / \ <--- make these two virtual to eliminate duplicate W in Z.
# X Y
# \ /
# Z
In your example:
class A{...};
class B : public virtual A{...};
class C : public virtual B{...}; // Edit: OP's code had this typo when I answered
class D : public B, public C{...};
Having B inherit virtually from A isn't necessary. The only virtual inheritances you need are C-->B and D-->B, since that is where the diamond "merges" going up the inheritance hierarchy:
# What you have | What you want?
# A | A
# / | /
# /v | /
# / | /
# B | B
# / \ | / \
# /v \ | /v \v
# / \ | / \
# C ) | C )
# \ / | \ /
# \ / | \ /
# \ / | \ /
# D | D
Of course if you have other classes not shown that inherit from A as well as B, that changes things -- maybe the B-->A inheritance does need to be virtual if there is another diamond you didn't tell us about.

Answer 4

I remember that in a non-virtual inheritance hierarchy, object of sub-class hold an object of its direct super-class.
That is not correct. Several implementations are going to do it this way, but it's not defined that way by Standard C++. Standard C++ does not specify how any of these things are to be implemented.
Virtual Inheritance is used only for some cases of multiple inheritance where a derived class multiply inherits from two base classes which themselves inherit from a common base class. An example of this is the iostream library, where istream and ostream inherit from basic_ios, and iostream inherits from istream and ostream (and so one iostream would have two basic_ios without virtual inheritance).
Unless you are in this specific scenario, you should not use virtual inheritance.
What about virtual inheritance? In this situation, does object of sub-class hold an object of its super-class directly or just hold a pointer pointing to an object of its super-class?
That is implementation defined. You do not need to know nor should you ever make any assumptions about this. Suffice to say that there is a runtime penalty for virtual inheritance, which is why you should avoid it when it is not needed.

Answer 5

Compare:
struct A {
void *vptr; // offset 0 size 4 alignment 4
char k[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
// MS:
struct base_B {
void *vptr; // offset 0 size 4 alignment 4
char j[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
A &a_subobject; // offset 8 size 4 alignment 4
// total size 12 alignment 4
base_B (&a_subobject) :a_subobject(a_subobject) {}
};
struct B {
base_B b; // offset 0 size 12 alignment 4
A a_subobject; // offset 12 size 8 alignment 4
// total size 20 alignment 4
B () : b(a_subobject) {}
};
struct base_C {
void *vptr; // offset 0 size 4 alignment 4
char i[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
A &a_subobject; // offset 8 size 4 alignment 4
// total size 12 alignment 4
base_C (&a_subobject) : a_subobject(a_subobject) {}
};
struct C {
base_C c;
A a_subobject; // offset 12 size 8 alignment 4
// total size 20 alignment 4
C () : c(a_subobject) {}
};
struct D {
// no new vptr!
// base_B is used as primary base: b_subobject.vptr is used as vptr
base_B b_subobject; // offset 0 size 12 alignment 4
base_C c_subobject; // offset 12 size 12 alignment 4
char h[3]; // offset 24 size 3 alignment 1
char unnamed_padding; // offset 27 size 1
A a_subobject; // offset 28 size 8 alignment 4
// total size 36 alignment 4
D (): b_subobject(a_subobject), c_subobject(a_subobject) {}
};
// GCC:
struct base_B {
void *vptr; // offset 0 size 4 alignment 4
char j[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
struct B {
base_B b; // offset 0 size 12 alignment 4
A a_subobject; // offset 8 size 8 alignment 4
// total size 16 alignment 4
};
struct base_C {
void *vptr; // offset 0 size 4 alignment 4
char i[3]; // offset 4 size 3 alignment 1
char unnamed_padding; // offset 7 size 1
// total size 8 alignment 4
};
struct C {
base_C b; // offset 0 size 12 alignment 4
A a_subobject; // offset 8 size 8 alignment 4
// total size 16 alignment 4
};
struct D {
// no new vptr!
// base_B is used as primary base: b_subobject.vptr is used as vptr
base_B b_subobject; // offset 0 size 8 alignment 4
base_C c_subobject; // offset 8 size 8 alignment 4
char h[3]; // offset 16 size 3 alignment 1
char unnamed_padding; // offset 19 size 1
A a_subobject; // offset 20 size 8 alignment 4
// total size 24 alignment 4
};