how to determine sizeof class with virtual functions?

问题

this is kind of homework question. For the following code,

#include <iostream>
using namespace std;

class A
{
public:
    virtual void f(){}
};

class B
{
public:
    virtual void f2(){}
};

class C: public A, public B
{
public: 
    virtual void f3(){}
};

class D: public C
{
public:
    virtual void f4(){}
};

int main()
{
    cout<<sizeof(D)<<endl;
}

The output is: 8

Could anyone please explain how it is 8 bytes? If the vtable implementation is compiler dependent, what should I answer for this kind of question in interviews? What about virtual base classes?

EDIT: i am working on a 32-bit platform.

回答1:

This is of course implementation-dependent. And it would make a terrible interview question. A good C++ programmer can just trust sizeof to be right and let the compiler worry about those vtable things.

But what's going on here is that a typical vtable-based implementation needs two vtables in objects of class C or D. Each base class needs its own vtable. The new virtual methods added by C and D can be handled by extending the vtable format from one base class, but the vtables used by A and B can't be combined.

In pseudo-C-code, here's how a most derived object of type D looks on my implementation (g++ 4.4.5 Linux x86):

void* D_vtable_part1[] = { (void*) 0, &D_typeinfo, &A::f1, &C::f3, &D::f4 };
void* D_vtable_part2[] = { (void*) -4, &D_typeinfo, &B::f2 };

struct D {
  void** vtable_A;
  void** vtable_B;
};

D d = { D_vtable_part1 + 1, D_vtable_part2 + 1 };

回答2:

In this question if you try to get Sizeof class A, it will give you answer '4' because A have only one virtual function so its __vptr will be of '4' byte.

In the same way, if you try to get Sizeof class B, it will give you answer '4' because B also have only one virtual function so its __vptr will be of '4' byte.

But class C is inheriting both classes A & B and C itself have a virtual function. So C will receive 2 __vptr pointers, and for its own virtual function C will use the inherited __vptr. So if you try to get Sizeof class C, it will give you answer '8' because C have two virtual pointers.

And lastly class D is inheriting class C so D will use inherited __vptr for its own virtual function and because class C have sizeof '8' byte so sizeof D will give you answer '8' byte.

回答3:

Forgive me for being vague, but you are mentioning that it is homework in nature.

See what sizeof() returns for the other classes. Your answer will vary depending on your compiler and whether you are in a 32 or 64 bit environment.

Happy sleuthing!

回答4:

The size of the object has nothing to do with how many methods that it has, nor with whether those methods are virtual or not. The size of an object is determined solely by its member variables.

I can't tell you exactly why you're getting a size of 8 bytes. Since there are no data members in your class, a C++ compiler could in principle generate a class that occupies no space at all [1]! I'm guessing that the 8 bytes is the minimum required to provide a pointer to the vtbl, plus possibly some padding.

[1] I think. No time to check the spec to find out at the moment whether sizeof can ever return 0.

来源：https://stackoverflow.com/questions/4766323/how-to-determine-sizeof-class-with-virtual-functions

标签

c++

sizeof

vtable