Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
203 views
in Technique[技术] by (71.8m points)

c++ - Object layout in case of virtual functions and multiple inheritance

I was recently asked in an interview about object layout with virtual functions and multiple inheritance involved.
I explained it in context of how it is implemented without multiple inheritance involved (i.e. how the compiler generated the virtual table, insert a secret pointer to the virtual table in each object and so on).
It seemed to me that there was something missing in my explanation.
So here are questions (see example below)

  1. What is the exact memory layout of the object of class C.
  2. Virtual tables entries for class C.
  3. Sizes (as returned by sizeof) of object of classes A, B and C. (8, 8, 16 ?? )
  4. What if virtual inheritance is used. Surely the sizes and virtual table entries should be affected ?

Example code:

class A {  
  public:   
    virtual int funA();     
  private:  
    int a;  
};

class B {  
  public:  
    virtual int funB();  
  private:  
    int b;  
};  

class C : public A, public B {  
  private:  
    int c;  
};   

Thanks!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

The memory layout and the vtable layout depend on your compiler. Using my gcc for instance, they look like this:

sizeof(int) == 4
sizeof(A) == 8
sizeof(B) == 8
sizeof(C) == 20

Note that sizeof(int) and the space needed for the vtable pointer can also vary from compiler to compiler and platform to platform. The reason why sizeof(C) == 20 and not 16 is that gcc gives it 8 bytes for the A subobject, 8 bytes for the B subobject and 4 bytes for its member int c.

Vtable for C
C::_ZTV1C: 6u entries
0     (int (*)(...))0
4     (int (*)(...))(& _ZTI1C)
8     A::funA
12    (int (*)(...))-0x00000000000000008
16    (int (*)(...))(& _ZTI1C)
20    B::funB

Class C
   size=20 align=4
   base size=20 base align=4
C (0x40bd5e00) 0
    vptr=((& C::_ZTV1C) + 8u)
  A (0x40bd6080) 0
      primary-for C (0x40bd5e00)
  B (0x40bd60c0) 8
      vptr=((& C::_ZTV1C) + 20u)

Using virtual inheritance

class C : public virtual A, public virtual B

the layout changes to

Vtable for C
C::_ZTV1C: 12u entries
0     16u
4     8u
8     (int (*)(...))0
12    (int (*)(...))(& _ZTI1C)
16    0u
20    (int (*)(...))-0x00000000000000008
24    (int (*)(...))(& _ZTI1C)
28    A::funA
32    0u
36    (int (*)(...))-0x00000000000000010
40    (int (*)(...))(& _ZTI1C)
44    B::funB

VTT for C
C::_ZTT1C: 3u entries
0     ((& C::_ZTV1C) + 16u)
4     ((& C::_ZTV1C) + 28u)
8     ((& C::_ZTV1C) + 44u)

Class C
   size=24 align=4
   base size=8 base align=4
C (0x40bd5e00) 0
    vptridx=0u vptr=((& C::_ZTV1C) + 16u)
  A (0x40bd6080) 8 virtual
      vptridx=4u vbaseoffset=-0x0000000000000000c vptr=((& C::_ZTV1C) + 28u)
  B (0x40bd60c0) 16 virtual
      vptridx=8u vbaseoffset=-0x00000000000000010 vptr=((& C::_ZTV1C) + 44u)

Using gcc, you can add -fdump-class-hierarchy to obtain this information.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...