C++ object size with virtual methods

Question

I have some questions about the object size with virtual.

1) virtual function

class A {
    public:
       int a;
       virtual voi         


        

Answer 1

This is all implementation defined. I'm using VC10 Beta2. The key to help understanding this stuff (the implementation of virtual functions), you need to know about a secret switch in the Visual Studio compiler, /d1reportSingleClassLayoutXXX. I'll get to that in a second.

The basic rule is the vtable needs to be located at offset 0 for any pointer to an object. This implies multiple vtables for multiple inheritance.

Couple questions here, I'll start at the top:

Does it mean that only one vptr is there even both of class B and class A have virtual function? Why there is only one vptr?

This is how virtual functions work, you want the base class and derived class to share the same vtable pointer (pointing to the implementation in the derived class.

It seems that in this case, two vptrs are in the layout.....How does this happen? I think the two vptrs one is for class A and another is for class B....so there is no vptr for the virtual function of class C?

This is the layout of class C, as reported by /d1reportSingleClassLayoutC:

class C size(20):
        +---
        | +--- (base class A)
 0      | | {vfptr}
 4      | | a
        | +---
        | +--- (base class B)
 8      | | {vfptr}
12      | | b
        | +---
16      | c
        +---

You are correct, there are two vtables, one for each base class. This is how it works in multiple inheritance; if the C* is casted to a B*, the pointer value gets adjusted by 8 bytes. A vtable still needs to be at offset 0 for virtual function calls to work.

The vtable in the above layout for class A is treated as class C's vtable (when called through a C*).

The sizeof B is 16 bytes -------------- Without virtual it should be 4 + 4 + 4 = 12. why there is another 4 bytes here? What's the layout of class B ?

This is the layout of class B in this example:

class B size(20):
        +---
 0      | {vfptr}
 4      | {vbptr}
 8      | b
        +---
        +--- (virtual base A)
12      | {vfptr}
16      | a
        +---

As you can see, there is an extra pointer to handle virtual inheritance. Virtual inheritance is complicated.

The sizeof D is 32 bytes -------------- it should be 16(class B) + 12(class C) + 4(int d) = 32. Is that right?

No, 36 bytes. Same deal with the virtual inheritance. Layout of D in this example:

class D size(36):
        +---
        | +--- (base class B)
 0      | | {vfptr}
 4      | | {vbptr}
 8      | | b
        | +---
        | +--- (base class C)
        | | +--- (base class A)
12      | | | {vfptr}
16      | | | a
        | | +---
20      | | c
        | +---
24      | d
        +---
        +--- (virtual base A)
28      | {vfptr}
32      | a
        +---

My question is , why there is an extra space when virtual inheritance is applied?

Virtual base class pointer, it's complicated. Base classes are "combined" in virtual inheritance. Instead of having a base class embedded into a class, the class will have a pointer to the base class object in the layout. If you have two base classes using virtual inheritance (the "diamond" class hierarchy), they will both point to the same virtual base class in the object, instead of having a separate copy of that base class.

What's the underneath rule for the object size in this case?

Important point; there are no rules: the compiler can do whatever it needs to do.

And a final detail; to make all these class layout diagrams I am compiling with:

cl test.cpp /d1reportSingleClassLayoutXXX

Where XXX is a substring match of the structs/classes you want to see the layout of. Using this you can explore the affects of various inheritance schemes yourself, as well as why/where padding is added, etc.

Answer 2

I am not sure but I think that it is because of pointer to Virtual method table

Answer 3

This obviously depends on the compiler implementation. Anyway I think that I can sum up the following rules from the implementation given by a classic paper linked below and which gives the number of bytes you get in your examples (except for class D which would be 36 bytes and not 32!!!):

The size of an object of class T is:

• The size of its fields PLUS the sum of the size of every object from which T inherits PLUS 4 bytes for every object from which T virtually inherits PLUS 4 bytes ONLY IF T needs ANOTHER v-table
• Pay attention: if a class K is virtually inherited multiple times (at any level) you have to add the size of K only once

So we have to answer another question: When does a class need ANOTHER v-table?

• A class that does not inherit from other classes needs a v-table only if it has one or more virtual methods
• OTHERWISE, a class needs another v-table ONLY IF NONE of the classes from which it non virtually inherits does have a v-table

The End of the rules (which I think can be applied to match what Terry Mahaffey has explained in his answer) :)

Anyway my suggestion is to read the following paper by Bjarne Stroustrup (the creator of C++) which explains exactly these things: how many virtual tables are needed with virtual or non virtual inheritance... and why!

It's really a good reading: http://www.hpc.unimelb.edu.au/nec/g1af05e/chap5.html

Answer 4

A good way to think about it is to understand what has to be done to handle up-casts. I'll try to answer your questions by showing the memory layout of objects of the classes you describe.

Code sample #2

The memory layout is as follows:

vptr | A::a | B::b

Upcasting a pointer to B to type A will result in the same address, with the same vptr being used. This is why there's no need for additional vptr's here.

Code sample #3

vptr | A::a | vptr | B::b | C::c

As you can see, there are two vptr's here, just like you guessed. Why? Because it's true that if we upcast from C to A we don't need to modify the address, and thus can use the same vptr. But if we upcast from C to B we do need that modification, and correspondingly we need a vptr at the start of the resulting object.

So, any inherited class beyond the first will require an additional vptr (unless that inherited class has no virtual methods, in which case it has no vptr).

Code sample #4 and beyond

When you derive virtually, you need a new pointer, called a base pointer, to point to the location in the memory layout of the derived classes. There can be more than one base pointer, of course.

So how does the memory layout look? That depends on the compiler. In your compiler it's probably something like

vptr | base pointer | B::b | vptr | A::a | C::c | vptr | A::a
          \-----------------------------------------^

But other compilers may incorporate base pointers in the virtual table (by using offsets - that deserves another question).

You need a base pointer because when you derive in a virtual fashion, the derived class will appear only once in the memory layout (it may appear additional times if it's also derived normally, as in your example), so all its children must point to the exact same location.

EDIT: clarification - it all really depends on the compiler, the memory layout I showed can be different in different compilers.

Answer 5

Quote> My question is, what's the rule about the number of vptrs in inheritance?

There are no rulez, every compiler vendor is allowed to implement the semantics of inheritance the way he sees fit.

class B: public A {}, size = 12. That's pretty normal, one vtable for B that has both virtual methods, vtable pointer + 2*int = 12

class C : public A, public B {}, size = 20. C can arbitrarily extend the vtable of either A or B. 2*vtable pointer + 3*int = 20

Virtual inheritance: that's where you really hit the edges of undocumented behavior. For example, in MSVC the #pragma vtordisp and /vd compile options become relevant. There's some background info in this article. I studied this a few times and decided the compile option acronym was representative for what could happen to my code if I ever used it.

Answer 6

All of this is completely implementation defined you realize. You can't count on any of it. There is no 'rule'.

In the inheritance example, here is how the virtual table for classes A and B might look:

      class A
+-----------------+
| pointer to A::v |
+-----------------+

      class B
+-----------------+
| pointer to A::v |
+-----------------+
| pointer to B::w |
+-----------------+

As you can see, if you have a pointer to class B's virtual table, it is also perfectly valid as class A's virtual table.

In your class C example, if you think about it, there is no way to make a virtual table that is both valid as a table for class C, class A, and class B. So the compiler makes two. One virtual table is valid for class A and C (mostly likely) and the other is valid for class A and B.