Understanding virtual inheritance class vtables and vptr creation

Question

The below code is of multiple inheritance where every class has one member variable, one normal function and one virtual function.

class basec
{
    int x;
public:
    basec()
    {
        x = 0;
    }
    void print()
    {}

    virtual void xyz()
    {}
};

class derivedc: public virtual basec
{
    int dc;
public:
    derivedc()
    {
        dc = 0;
    }
    virtual void xyzdc()
    {}
};

class  derivedd: public virtual basec
{
    int dd;
public:
    derivedd()
    {
        dd = 0;
    }
    virtual void xyzdd()
    {}
};

class child: public derivedc, public derivedd
{
    char de;
public:
    child()
    {
        de = '4';
    }
    virtual void xyzde()
    {}
};

main(int argc, char **argv)
{
    basec bobj, bobjc;    
    derivedc dcobj;
    derivedd ddobj;
    child deobj;

    std::cout << "C " << sizeof(basec) << endl;
    std::cout << "D " << sizeof(derivedc) << endl;
    std::cout << "E " << sizeof(derivedd) << endl;
    std::cout << "F " << sizeof(child) << endl;

    return(0);
}

Output of the program is as follows:

main()
C 8
D 16
E 16
F 28

Seeing each of the object in gdb I see below:

(gdb) p/x bobj
$1 = {_vptr.basec = 0x8048c80, x = 0x0}

(gdb) p/x dcobj
$3 = {<basec> = {_vptr.basec = 0x8048c5c, x = 0x0}, _vptr.derivedc = 0x8048c4c, dc = 0x0}

(gdb) p/x ddobj
$4 = {<basec> = {_vptr.basec = 0x8048c1c, x = 0x0}, _vptr.derivedd = 0x8048c0c, dd = 0x0}

(gdb) p/x deobj
$5 = {<derivedc> = {<basec> = {_vptr.basec = 0x8048b90, x = 0x0}, _vptr.derivedc = 0x8048b6c, dc = 0x0}, <derivedd> = {_vptr.derivedd = 0x8048b80, dd = 0x0}, de = 0x34}

We see that in the base class "basec" and in each of the virtually derived classes "derivedc" and "derivedd" objects a vptr is added for their vtables.

Question is why the child class inspite of having a virtual function does not have its own vtable and why no vptr of its own in its object? In which class's vtable will the virtual function of child class appear?

Perhaps the compiler decided it didn't need a vtable, since there are no override methods anywhere in your example. What happens if you name each of the virtual functions to just xyz? — Tim Randall
– Tim Randall, Commented Sep 20, 2018 at 18:38
Thanks Tim for the reply. Will try your suggestion and check. — RKum
– RKum, Commented Sep 20, 2018 at 19:18
For example clang 7 put the child::xyzde() in the derivedc vtable in child see "vtable for child" - gcc.godbolt.org/z/d2VWza. If you change child to class child: public virtual derivedc, public virtual derivedd you get three separate tables. gcc.godbolt.org/z/wCbmWb. An if you remove all virtual inheritance you get basec copied as expected - gcc.godbolt.org/z/8wxPGA. — Mihayl
– Mihayl, Commented Sep 20, 2018 at 19:30
vtable for child:... .quad derivedc::xyzdc() .quad child::xyzde() .... .quad typeinfo for child .quad derivedd::xyzdd() ... .quad typeinfo for child .quad basec::xyz() — Mihayl
– Mihayl, Commented Sep 20, 2018 at 19:37
Maybe you need to experiment with very simple inheritance cases and see how many vptr are used and what the vtable contain before trying to understand that "diamond". — curiousguy
– curiousguy, Commented Sep 23, 2018 at 1:51

Mihayl · Accepted Answer · 2018-10-15 07:56:02Z

The compiler is free to put it in one on of the existing vtables, just like normal inheritance works. The virtual inheritance guarantees you that you have the virtual base class only once.

For example both clang 7 and gcc 8.2 put the child::xyzde() in the derivedc's vtable in child. See "vtable for child" (clang 7 and gcc 8.2 on godbolt).

class derivedc : public virtual basec
class derivedd : public virtual basec
class child: public derivedc, public derivedd

vtable for child:
 .quad 32
 .quad 0
 .quad typeinfo for child
 .quad derivedc::xyzdc()
 .quad child::xyzde()       <- child::xyzde() together with derivedc's methods
 .quad 16
 .quad -16
 .quad typeinfo for child
 .quad derivedd::xyzdd()
 .quad 0
 .quad -32
 .quad typeinfo for child
 .quad basec::xyz()         <- basec is only once in child

If you change child base classes to virtual as below you get three separate tables:

class child: public virtual derivedc, public virtual derivedd

clang on godbolt:

vtable for child:
 .quad 48
 .quad 32
 .quad 16
 .quad 0
 .quad typeinfo for child
 .quad child::xyzde()      <- New vtable for child
 .quad 0
 .quad 16
 .quad -16
 .quad typeinfo for child
 .quad derivedc::xyzdc()
 .quad 0
 .quad -32
 .quad typeinfo for child
 .quad basec::xyz()        <- basec is only once in child
 .quad 0
 .quad -16
 .quad -48
 .quad typeinfo for child
 .quad derivedd::xyzdd()

And if you remove all virtual inheritance you get basec twice in child as expected (clang on godbolt).

class derivedc : public basec
class derivedd : public basec
class child: public derivedc, public derivedd

vtable for child:
 .quad 0
 .quad typeinfo for child
 .quad basec::xyz()        <- basec from derivedc
 .quad derivedc::xyzdc()
 .quad child::xyzde()      <- child::xyzde() together with derivedc's methods
 .quad -16
 .quad typeinfo for child
 .quad basec::xyz()        <- basec from derivedd
 .quad derivedd::xyzdd()

C++ vtables - Part 3 - Virtual Inheritance has some short explanation of VTT and construction vtable for X-in-child.

C++ vtables - Part 3 - Virtual Inheritance gives a good explanation.

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