vptr help? [modified]

sam_psycho

Please tell me how function call is resolve, how vptr works in this case CODE:

class base
{
public:
virtual void f1()
{
cout<<"base f1";
}
virtual void f2()
{
cout<<"base f2";
}
};
class d1: public base
{
public:
virtual void f1()
{
cout<<"d1's f1";
}

virtual void f2()
{
cout<<"d1's f2";
}
};
class d2: virtual public base,virtual public d1
{
public:

d2()
{
  cout<<"d2 class construxtor\\n";
}
~d2()
{
  cout<<" d2 class destructor\\n";
}

};

void main()
{
clrscr();
base* obj= new d2;
obj->f1();
obj->f2();
}

modified on Sunday, July 12, 2009 2:46 PM

Stuart Dootson · modified on Sunday, July 12, 2009 2:46 PM

vptrs don't come into it - you have no virtual functions, so therefore there will be no vtable. If the compiler was able to construct your base pointer obj (it can't because it inherits from base twice - which one should it upcast through?), there would be no runtime resolution of f1 and f2; they're non-virtual functions, so they will be resolved using the static type of obj, i.e. base. That means that the program (if it could compile) would output base f1base f2

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Graham Shanks

Actually the code will (and does) compile. Note that base is inherited virtually (note the use of virtual public base in the declaration of class d2). This ensures that only one version of the base class will be included in d2 However, you are completely correct in that there is no vtable, and the output wil be base f1base f2.

Graham Librarians rule, Ook!

sam_psycho

all right I got it...but if I am making f1() and f2() virtual in base class and d1 class, how vptr will work?

Stuart Dootson

Better tell that to gcc, then - it gives this warning and error:

a.cpp:30: warning: direct base ‘base’ inaccessible in ‘d2’ due to ambiguity
a.cpp: In function ‘int main()’:
a.cpp:46: error: ‘base’ is an ambiguous base of ‘d2’

VC++ 2008 agrees

a.cpp(30) : error C2584: 'd2' : direct base 'base' is inaccessible; already a base of 'd1'
a.cpp(5) : see declaration of 'base'
a.cpp(16) : see declaration of 'd1'

The virtual qualifier needs to be applied to all inheritance instances of the shared base, so you need this code for it to compile (note virtual wherever 'base' is inherited):

class base
{
public:
void f1()
{
cout<<"base f1";
}
void f2()
{
cout<<"base f2";
}
};
class d1: virtual public base
{
public:
void f1()
{
cout<<"d1's f1";
}

void f2()
{
cout<<"d1's f2";
}
};
class d2: virtual public base, public d1
{
public:

d2()
{
  cout<<"d2 class construxtor\\n";
}
~d2()
{
  cout<<" d2 class destructor\\n";
}

};

int main()
{
base* obj= new d2;
obj->f1();
obj->f2();
}

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Graham Shanks

OK, you caught me. :-O That'll learn me not to trust VS2005 :doh:

Graham Librarians rule, Ook!

Stuart Dootson

As base is declared virtual, there is one instance of base in d2. This means (IIUC) that the functions in d1 will hide the functions in base from the point-of-view of d2 because d1 is more derived than base.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

Stuart Dootson

Graham Shanks wrote:

VS2005

That compiled in VS2005? I guess I can understand why, I think. Maybe. Also yet another example why it's good to keep clear of tricksy inheritance features in C++. a) the programmer isn't 100% sure what should happen (I'm definitely including myself in that group), and b) neither's the compiler.

Java, Basic, who cares - it's all a bunch of tree-hugging hippy cr*p

_Superman_

For the working of vptr, allow me to redirect you to one of my articles Polymorphism in C[^]

«_Superman_» I love work. It gives me something to do between weekends.

Joe Woodbury

Stuart Dootson wrote:

That compiled in VS2005?

No. Doesn't compile in VS 2005, 2008 or VC++ 6.0.

Joe Woodbury

The code doesn't compile with VC++ 6.0, VS 2005 and VS 2008. (clrscr() is also undeclared in Visual C++; I vaguely recall it being in Turbo C. Perhaps he has an early version of Turbo C++ and it's not showing this as an error.)