Object layout in case of virtual functions and multiple inheritance

Solution 1:

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.

Solution 2:

1 thing to expect with multiple inheritance is that your pointer can change when casting to a (typically not first) subclass. Something you should be aware of while debugging and answering interview questions.

Solution 3:

First, a polymorphic class has at least one virtual function, so it has a vptr:

struct A {
    virtual void foo();
};

is compiled to:

struct A__vtable { // vtable for objects of declared type A
    void (*foo__ptr) (A *__this); // pointer to foo() virtual function
};

void A__foo (A *__this); // A::foo ()

// vtable for objects of real (dynamic) type A
const A__vtable A__real = { // vtable is never modified
    /*foo__ptr =*/ A__foo
};

struct A {
    A__vtable const *__vptr; // ptr to const not const ptr
                             // vptr is modified at runtime
};

// default constructor for class A (implicitly declared)
void A__ctor (A *__that) { 
    __that->__vptr = &A__real;
}

Remark: C++ can be compiled to another high level language like C (as cfront did) or even to a C++ subset (here C++ without virtual). I put __ in compiler generated names.

Note that this is a simplistic model where RTTI is not supported; real compilers will add data in the vtable to support typeid.

Now, a simple derived class:

struct Der : A {
    override void foo();
    virtual void bar();
};

Non-virtual (*) base class subobjects are subobjects like member subobjects, but while member subobjects are complete objects, ie. their real (dynamic) type is their declared type, base class subobjects are not complete, and their real type change during construction.

(*) virtual bases are very different, like virtual member functions are different from non virtual members

struct Der__vtable { // vtable for objects of declared type Der
    A__vtable __primary_base; // first position
    void (*bar__ptr) (Der *__this); 
};

// overriding of a virtual function in A:
void Der__foo (A *__this); // Der::foo ()

// new virtual function in Der:
void Der__bar (Der *__this); // Der::bar ()

// vtable for objects of real (dynamic) type Der
const Der__vtable Der__real = { 
    { /*foo__ptr =*/ Der__foo },
    /*foo__ptr =*/ Der__bar
};

struct Der { // no additional vptr
    A __primary_base; // first position
};

Here "first position" means that the member must be first (other members could be re-ordered): they are located at offset zero so we can reinterpret_cast pointers, the types are compatibles; at non-zero offset, we would have to do pointer adjustments with arithmetic on char*.

The lack of adjustment may not seem like a big deal in term of generated code (just some add immediate asm instructions), but it means much more than that, it means that such pointers can be viewed as having different types: an object of type A__vtable* can contain a pointer to Der__vtable and be treated as either a Der__vtable* or a A__vtable*. The same pointer object serves as a pointer to a A__vtable in functions dealing with objects of type A and as a pointer to a Der__vtable in functions dealing with objects of type Der.

// default constructor for class Der (implicitly declared)
void Der__ctor (Der *__this) { 
    A__ctor (reinterpret_cast<A*> (__this));
    __this->__vptr = reinterpret_cast<A__vtable const*> (&Der__real);
}

You see that the dynamic type, as defined by the vptr, changes during construction as we assign a new value to the vptr (in this particular case the call to the base class constructor does nothing useful and can be optimised away, but it isn't the case with non trivial constructors).

With multiple inheritance:

struct C : A, B {};

A C instance will contain a A and a B, like that:

struct C {
    A base__A; // primary base
    B base__B;
};

Note that only one of these base class subobjects can have the privilege of sitting at offset zero; this is important in many ways:

conversion of pointers to other base classes (upcasts) will need an adjustment; conversely, upcasts need the opposite adjustments;
this implies that when doing a virtual call with a base class pointer, the this has the correct value for entry in the derived class overrider.

So the following code:

void B::printaddr() {
    printf ("%p", this);
}

void C::printaddr () { // overrides B::printaddr()
    printf ("%p", this);
}

can be compiled to

void B__printaddr (B *__this) {
    printf ("%p", __this);
}

// proper C::printaddr taking a this of type C* (new vtable entry in C)
void C__printaddr (C *__this) {
    printf ("%p", __this);
}

// C::printaddr overrider for B::printaddr
// needed for compatibility in vtable
void C__B__printaddr (B *__this) {
    C__printaddr (reinterpret_cast<C*>(reinterpret_cast<char*> (__this) - offset__C__B));
}

We see the C__B__printaddr declared type and semantics are compatibles with B__printaddr, so we can use &C__B__printaddr in the vtable of B; C__printaddr is not compatible but can be used for calls involving a C objects, or classes derived from C.

A non virtual member function is like a free function that has access to internal stuff. A virtual member function is "flexibility point" which can be customised by overriding. virtual member function declaration play a special role in the definition of a class: like other members they are part of the contract with the external world, but at the same time they are part of a contract with derived class.

A non virtual base class is like a member object where we can refine behaviour via overriding (also we can access protected members). For the external world, the inheritance for A in Der implies that implicit derived-to-base conversions will exist for pointers, that a A& can be bound to a Der lvalue, etc. For further derived classes (derived from Der), it also means that virtual functions of A are inherited in the Der: virtual functions in A can be overridden in further derived classes.

When a class is further derived, say Der2 is derived from Der, implicit conversions a pointers of type Der2* to A* is semantically performed in step: first, a conversion to Der* is validated (the access control to the inheritance relation of Der2 from Der is checked with the usual public/protected/private/friend rules), then the access control of Der to A. A non virtual inheritance relation cannot be refined or overridden in derived classes.

Non virtual members functions can be directly called and virtual members must be called indirectly via the vtable (unless the real object type happens to be known by the compiler), so the virtual keyword adds an indirection to members functions access. Just like for function members, the virtual keyword adds an indirection to base object access; just like for functions, virtual base classes add a flexibility point in inheritance.

When doing non-virtual, repeated, multiple inheritance :

struct Top { int i; };
struct Left : Top { };
struct Right : Top { };
struct Bottom : Left, Right { };

There are just two Top::i subobjects in Bottom (Left::i and Right::i), as with members objects:

struct Top { int i; };
struct mLeft { Top t; };
struct mRight { mTop t; };
struct mBottom { mLeft l; mRight r; }

Nobody is surprised that there are two int sub-members (l.t.i and r.t.i).

With virtual functions:

struct Top { virtual void foo(); };
struct Left : Top { }; // could override foo
struct Right : Top { }; // could override foo
struct Bottom : Left, Right { }; // could override foo (both)

it means there are two different (unrelated) virtual functions called foo, with distinct vtable entries (both as they have the same signature, they can have a common overrider).

The semantic of non virtual base classes follows from the fact that basic, non virtual, inheritance is an exclusive relation: the inheritance relation established between Left and Top cannot be modified by a further derivation, so the fact that a similar relation exists between Right and Top cannot affect this relation. In particular, it means that Left::Top::foo() can be overridden in Left and in Bottom, but Right, which has no inheritance relation with Left::Top, cannot set this customisation point.

Virtual base classes are different: a virtual inheritance is a shared relation that can be customised in derived classes:

struct Top { int i; virtual void foo(); };
struct vLeft : virtual Top { }; 
struct vRight : virtual Top { };
struct vBottom : vLeft, vRight { };

Here, this is only one base class subobject Top, only one int member.

Implementation:

Room for non virtual base classes are allocated based on a static layout with fixed offsets in the derived class. Note that the layout of a derived class is the included in the layout of more derived class, so the exact position of subobjects does not depend on the real (dynamic) type of object (just like the address of a non virtual function is a constant). OTOH, the position of subobjects in a class with virtual inheritance is determined by the dynamic type (just like the address of the implementation of a virtual function is known only when the dynamic type is known).

The location of subobject will be determined at runtime with the vptr and the vtable (reuse of the existing vptr implies less space overhead), or a direct internal pointer to the subobject (more overhead, less indirections needed).

Because the offset of a virtual base class is determined only for a complete object, and cannot be known for a given declared type, a virtual base cannot be allocated at offset zero and is never a primary base. A derived class will never reuse the vptr of a virtual base as its own vptr.

In term of possible translation:

struct vLeft__vtable { 
    int Top__offset; // relative vLeft-Top offset
    void (*foo__ptr) (vLeft *__this); 
    // additional virtual member function go here
};

// this is what a subobject of type vLeft looks like
struct vLeft__subobject { 
    vLeft__vtable const *__vptr;
    // data members go here
};

void vLeft__subobject__ctor (vLeft__subobject *__this) { 
    // initialise data members
}

// this is a complete object of type vLeft 
struct vLeft__complete {
    vLeft__subobject __sub;
    Top Top__base;
}; 

// non virtual calls to vLeft::foo
void vLeft__real__foo (vLeft__complete *__this);

// virtual function implementation: call via base class
// layout is vLeft__complete 
void Top__in__vLeft__foo (Top *__this) {
    // inverse .Top__base member access 
    char *cp = reinterpret_cast<char*> (__this);
    cp -= offsetof (vLeft__complete,Top__base);
    vLeft__complete *__real = reinterpret_cast<vLeft__complete*> (cp);
    vLeft__real__foo (__real);
}

void vLeft__foo (vLeft *__this) {
    vLeft__real__foo (reinterpret_cast<vLeft__complete*> (__this));
}

// Top vtable for objects of real type vLeft
const Top__vtable Top__in__vLeft__real = { 
    /*foo__ptr =*/ Top__in__vLeft__foo 
};

// vLeft vtable for objects of real type vLeft
const vLeft__vtable vLeft__real = { 
    /*Top__offset=*/ offsetof(vLeft__complete, Top__base),
    /*foo__ptr =*/ vLeft__foo 
};

void vLeft__complete__ctor (vLeft__complete *__this) { 
    // construct virtual bases first
    Top__ctor (&__this->Top__base); 

    // construct non virtual bases: 
    // change dynamic type to vLeft
    // adjust both virtual base class vptr and current vptr
    __this->Top__base.__vptr = &Top__in__vLeft__real;
    __this->__vptr = &vLeft__real;

    vLeft__subobject__ctor (&__this->__sub);
}

For an object of known type, access to the base class is through vLeft__complete:

struct a_vLeft {
    vLeft m;
};

void f(a_vLeft &r) {
    Top &t = r.m; // upcast
    printf ("%p", &t);
}

is translated to:

struct a_vLeft {
    vLeft__complete m;
};

void f(a_vLeft &r) {
    Top &t = r.m.Top__base;
    printf ("%p", &t);
}

Here the real (dynamic) type of r.m is known and so is the relative position of the subobject is known at compile time. But here:

void f(vLeft &r) {
    Top &t = r; // upcast
    printf ("%p", &t);
}

the real (dynamic) type of r is not known, so the access is through the vptr:

void f(vLeft &r) {
    int off = r.__vptr->Top__offset;
    char *p = reinterpret_cast<char*> (&r) + off;
    printf ("%p", p);
}

This function can accept any derived class with a different layout:

// this is what a subobject of type vBottom looks like
struct vBottom__subobject { 
    vLeft__subobject vLeft__base; // primary base
    vRight__subobject vRight__base; 
    // data members go here
};

// this is a complete object of type vBottom 
struct vBottom__complete {
    vBottom__subobject __sub; 
    // virtual base classes follow:
    Top Top__base;
};

Note that the vLeft base class is at a fixed location in a vBottom__subobject, so vBottom__subobject.__ptr is used as a vptr for the whole vBottom.

Semantics:

The inheritance relation is shared by all derived classes; this means that the right to override is shared, so vRight can override vLeft::foo. This creates a sharing of responsibilities: vLeft and vRight must agree on how they customise Top:

struct Top { virtual void foo(); };
struct vLeft : virtual Top { 
    override void foo(); // I want to customise Top
}; 
struct vRight : virtual Top { 
    override void foo(); // I want to customise Top
}; 
struct vBottom : vLeft, vRight { };  // error

Here we see a conflict: vLeft and vRight seek to define the behaviour of the only foo virtual function, and vBottom definition is in error for lack of a common overrider.

struct vBottom : vLeft, vRight  { 
    override void foo(); // reconcile vLeft and vRight 
                         // with a common overrider
};

Implementation:

The construction of class with non virtual base classes with non virtual base classes involves calling base class constructors in the same order as done for member variables, changing the dynamic type each time we enter a ctor. During construction, the base class subobjects really act as if they were complete objects (this is even true with impossible complete abstract base class subobjects: they are objects with undefined (pure) virtual functions). Virtual functions and RTTI can be called during construction (except of course pure virtual functions).

The construction of a class with non virtual base classes with virtual bases is more complicated: during construction, the dynamic type is the base class type, but the layout of virtual base is still the layout of the most derived type that is not yet constructed, so we need more vtables to describe this state:

// vtable for construction of vLeft subobject of future type vBottom
const vLeft__vtable vLeft__ctor__vBottom = { 
    /*Top__offset=*/ offsetof(vBottom__complete, Top__base),
    /*foo__ptr =*/ vLeft__foo 
};

The virtual functions are those of vLeft (during construction, the vBottom object lifetime has not begun), while the virtual base locations are those of a vBottom (as defined in the vBottom__complete translated objected).

Semantics:

During initialisation, it is obvious that we must be careful not to use an object before it is initialised. Because C++ gives us a name before an object is fully initialised, it is easy to do that:

int foo (int *p) { return *pi; }
int i = foo(&i);

or with the this pointer in the constructor:

struct silly { 
    int i;
    std::string s;
    static int foo (bad *p) { 
        p->s.empty(); // s is not even constructed!
        return p->i; // i is not set!
    }
    silly () : i(foo(this)) { }
};

It is pretty obvious that any use of this in the ctor-init-list must be carefully checked. After initialisation of all members, this can be passed to other functions and registered in some set (until destruction begins).

What is less obvious is that when construction of a class involving shared virtual bases, subobjects stop being constructed: during the construction of a vBottom:

first the virtual bases are constructed: when Top is constructed, it is constructed like a normal subject (Top doesn't even know it is virtual base)
then the base classes are constructed in left to right order: the vLeft subobject is constructed and becomes functional as a normal vLeft (but with a vBottom layout), so the Top base class subobject now has a vLeft dynamic type;
the vRight subobject construction begins, and the dynamic type of the base class changes to vRight; but vRight isn't derived from vLeft, doesn't know anything about vLeft, so the vLeft base is now broken;
when the body of the Bottom constructor begins, the types of all subobjects have stabilised and vLeft is functional again.

Solution 4:

I am not sure how this answer can be taken as a complete answer without the mention of alignment or padding bits.

Let me give a bit background of Alignment:

"A memory address a, is said to be n-byte aligned when a is a multiple of n bytes (where n is a power of 2). In this context a byte is the smallest unit of memory access, i.e. each memory address specifies a different byte. An n-byte aligned address would have log2(n) least-significant zeros when expressed in binary.

The alternate wording b-bit aligned designates a b/8 byte aligned address (ex. 64-bit aligned is 8 bytes aligned).

A memory access is said to be aligned when the datum being accessed is n bytes long and the datum address is n-byte aligned. When a memory access is not aligned, it is said to be misaligned. Note that by definition byte memory accesses are always aligned.

A memory pointer that refers to primitive data that is n bytes long is said to be aligned if it is only allowed to contain addresses that are n-byte aligned, otherwise it is said to be unaligned. A memory pointer that refers to a data aggregate (a data structure or array) is aligned if (and only if) each primitive datum in the aggregate is aligned.

Note that the definitions above assume that each primitive datum is a power of two bytes long. When this is not the case (as with 80-bit floating-point on x86) the context influences the conditions where the datum is considered aligned or not.

Data structures can be stored in memory on the stack with a static size known as bounded or on the heap with a dynamic size known as unbounded." - from Wiki...

To maintain the alignment, compiler inserts padding bits in the compiled code of a structure/class object. " Although the compiler (or interpreter) normally allocates individual data items on aligned boundaries, data structures often have members with different alignment requirements. To maintain proper alignment the translator normally inserts additional unnamed data members so that each member is properly aligned. In addition the data structure as a whole may be padded with a final unnamed member. This allows each member of an array of structures to be properly aligned. .... ....

Padding is only inserted when a structure member is followed by a member with a larger alignment requirement or at the end of the structure" - Wiki

To get more info about how GCC does it, please look at

http://www.delorie.com/gnu/docs/gcc/gccint_111.html

and search for the text "basic-align"

Now lets come to this problem:

Using the example class, i have created this program for a GCC compiler running on a 64 bit Ubuntu.

int main() {
    cout << "!!!Hello World!!!" << endl; // prints !!!Hello World!!!
    A objA;
    C objC;
    cout<<__alignof__(objA.a)<<endl;
    cout<<sizeof(void*)<<endl;
    cout<<sizeof(int)<<endl;
    cout<<sizeof(A)<<endl;
    cout<<sizeof(B)<<endl;
    cout<<sizeof(C)<<endl;
    cout<<__alignof__(objC.a)<<endl;
    cout<<__alignof__(A)<<endl;
    cout<<__alignof__(C)<<endl;
    return 0;
}

And the result for this program is as the following:

Now let me explain it. As both A & B have virtual functions, they will create seperate VTABLEs and VPTR will be added at the beginning of their objects, respectively.

Hence object of class A will have a VPTR (pointing to the VTABLE of A) and an int. The pointer will be 8 byte long and the int will be 4 byte long. Hence before compile the size is 12 bytes. But the compiler will add extra 4 bytes at the end of int a as padding bits. Hence after compilation, the A's objects size will be 12+4 = 16.

Similarly for class B's objects.

Now object of C will have two VPTRs (one for each class A & class B) and 3 ints (a,b,c). So the size should have been 8 (VPTR A) + 4 (int a) + 4 (padding bytes) + 8 (VPTR B) + 4 (int b) + 4 (int c) = 32 bytes. So the total size of C will be 32 bytes.

Object layout in case of virtual functions and multiple inheritance

Solution 1:

Solution 2:

Solution 3:

Solution 4:

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