Virtual function and multiple inheritance

Hello everyone,


In the following multiple inheritance sample code, I have tested in class Derived, there are two __vfptr, pointing to the virtual function table for Foo and Goo repectively -- i.e. 8 bytes, 2 pointer on 32-bit machine.

My questions,

1. Why two __vfptr is needed? Why not just one?
2. class Derived has its own virtual method func2, why it does not have its own virtual function table pointer?

Code:

class Foo {
public:
	virtual int func() {return 1;};
};

class Goo {
public:
	virtual int func() {return 2;};
};

class Derived: Foo, Goo {
public:
	virtual int func2() {return 3;};
	int increase() {return -1;};
	int decrease() {return -2;};
};

int main()
{
	Derived d;
	int size;

	size = sizeof (d); // size is 8, two __vfptr?

	return 0;
}

thanks in advance,
George

Foo and Goo are unrelated classes with virtual functions. Thus, they both need a vptr.

When combining them to be part of Derived, they can't change layout. It would break code. Thus, the combined Derived object has both vptrs.

The Derived object already has two vptrs to choose from. Why would it add yet another?

Thanks CornedBee,


I agree with your theory, but it does not cover one of my confusions, which is, where is class Derived's own __vfptr (points to its own virtual methods)? From debugging, I can not find it out by adding a new method in Derived which is not in Foo and Goo. Like, we add a new virtual method called func_derived.

Code:

virtual int func_derived() {return 4;}

regards,
George

Originally Posted by CornedBee

Foo and Goo are unrelated classes with virtual functions. Thus, they both need a vptr.

When combining them to be part of Derived, they can't change layout. It would break code. Thus, the combined Derived object has both vptrs.

The Derived object already has two vptrs to choose from. Why would it add yet another?

I repeat:

The Derived object already has two vptrs to choose from. Why would it add yet another?

See, the way virtual tables work, there is one for every class. Derived doesn't need another vptr - it simply lets its base class vptrs point to its own table.

But here is an interesting question: what if the class has two base classes? Which vptr will it use in this case? I'll admit, I've never studied the vptr table.

That's one of the details in the vptr implementation that may very well differ between compilers.

But typically, I'd say it uses that of the first base class.

Hi CornedBee,


I have posted my code and screen snapshot, you can see in Derived class there are two vfptr pointing two different virtual function table.

See the code and attachment, thanks.

Any ideas?

Code:

class Foo {
public:
	virtual int func_foo() {return 1;};
};

class Goo {
public:
	virtual int func_goo() = 0;
};

class Goo2: public Goo {
public:
	virtual int func_goo() {return 2;};
};

class Goo3: public Goo2 {
public:
	virtual int func_goo() {return 2;};
};

class Derived: Foo, Goo {
public:
	virtual int func_foo() {return 3;};
	virtual int func_goo() {return 2;};
	virtual int func_derived() {return 4;}
	int increase() {return -1;};
	int decrease() {return -2;};
};

int main()
{
	Derived d;
	Goo2 g2;
	Goo3 g3;
	int size;

	size = sizeof (d); // size is 8, two __vfptr?
	size = sizeof (g2);
	size = sizeof (g3);

	return 0;
}

Originally Posted by CornedBee

I repeat:


See, the way virtual tables work, there is one for every class. Derived doesn't need another vptr - it simply lets its base class vptrs point to its own table.

regards,
George

Hi Elysia,


You can find detailed information in my post #7 when there are two vfptr in one class. I provide code and screen snapshots. It is appreciated if you could add some ideas. :-)

Originally Posted by Elysia

But here is an interesting question: what if the class has two base classes? Which vptr will it use in this case? I'll admit, I've never studied the vptr table.

regards,
George

CornedBee already told you: Derived inherits the vtable pointers from its base classes!
The debug window shows it as well. Those pointers are part of the Foo and Goo base classes.
What more do you need to know?

Derived::vftable for Goo
Derived::vftable for Foo

How much clearer can the debugger be?



... not that it matters. That stuff is interesting if you want to implement a C++ compiler yourself.

Thanks Elysia,


Why virtual function func_derived of class Derived is not in virtual table of class Derived? In virtual function theory, all virtual tables should be in vtable.

Originally Posted by Elysia

CornedBee already told you: Derived inherits the vtable pointers from its base classes!
The debug window shows it as well. Those pointers are part of the Foo and Goo base classes.
What more do you need to know?

regards,
George

Hi CornedBee,


This is what you mentioned before.

"Derived doesn't need another vptr - it simply lets its base class vptrs point to its own table."

But if you debug my simlpe program, Derived class's own virtual function func_derived is not in vtable. Why?

Originally Posted by CornedBee

Derived::vftable for Goo
Derived::vftable for Foo

How much clearer can the debugger be?



... not that it matters. That stuff is interesting if you want to implement a C++ compiler yourself.

regards,
George

Originally Posted by George2

Thanks Elysia,

Why virtual function func_derived of class Derived is not in virtual table of class Derived? In virtual function theory, all virtual tables should be in vtable.

regards,
George

Take a look into the vptr of the base classes and see if you can find it there.

Thanks Elysia,


I can not find virtual method in vtable func_derived and it is missed in vtable of class Derived. Can you find it or meet with similar issues before -- not all virtual methods are in vtable?

Originally Posted by Elysia

Take a look into the vptr of the base classes and see if you can find it there.

regards,
George

I have no idea how it stores the functions the vtables (in case of Derived). They certainly don't show up in the debugger, but they certainly work, so they must be in a vtable. In fact, the assembly shows this.