Size of an inherited class ?

[François Fleuret]

Dear all,

I fear this is a question which has been asked zillions of
times. However, it seems to me to be such a legitimate feature that I
can't imagine it is impossible.

How can one know the size of an object in a "virtual" manner ?

For instance, how can I write a virtual function print_size(); which
prints the size of *this, and which will print the correct value with
an object of an inherited class ?

Thanks in advance for any hint,

--
François Fleuret http://cvlab.epfl.ch/~fleuret

[Michiel.Salters]

François Fleuret wrote:

Dear all,

I fear this is a question which has been asked zillions of
times. However, it seems to me to be such a legitimate feature that I
can't imagine it is impossible.

How can one know the size of an object in a "virtual" manner ?

For instance, how can I write a virtual function print_size(); which
prints the size of *this, and which will print the correct value with
an object of an inherited class ?

You can't, because you can't do anything useful with it. sizeof() is
used to allocate memory for new objects (before you can even call
print_size()) or memcpy existing objects (which you can't do if it has
virtual functions).

What legitimate use did you imagine?

HTH,
Michiel Salters

[John Carson]

"François Fleuret" <fr**************@epfl.ch> wrote in message
news:87************@fleuret.homeunix.org

Dear all,

I fear this is a question which has been asked zillions of
times. However, it seems to me to be such a legitimate feature that I
can't imagine it is impossible.

How can one know the size of an object in a "virtual" manner ?

For instance, how can I write a virtual function print_size(); which
prints the size of *this, and which will print the correct value with
an object of an inherited class ?

Thanks in advance for any hint,

--
François Fleuret
http://cvlab.epfl.ch/~fleuret

This seems to work

#include <iostream>
using namespace std;

struct Base
{
int b;
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};
struct Derived : Base
{
int array[4];
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};

int main(int argc, char* argv[])
{
Base b;
Derived d;

Base * pb1 = &b;
Base * pb2 = &d;

pb1->Print_Size();
pb2->Print_Size();
return 0;
}
--
John Carson

[peter koch]

Mi*************@tomtom.com skrev:

François Fleuret wrote:

Dear all,

I fear this is a question which has been asked zillions of
times. However, it seems to me to be such a legitimate feature that I
can't imagine it is impossible.

How can one know the size of an object in a "virtual" manner ?

For instance, how can I write a virtual function print_size(); which
prints the size of *this, and which will print the correct value with
an object of an inherited class ?
You can't, because you can't do anything useful with it. sizeof() is
used to allocate memory for new objects (before you can even call
print_size()) or memcpy existing objects (which you can't do if it has
virtual functions).

What legitimate use did you imagine?

It is difficult to find a use for that size, but perhaps some sort of
clone:

class base
{
public:
virtual ~base() = 0;
virtual int size() = 0;
virtual void clone(void* adress) = 0;
};

void silly(base* b)
{
void* clonemem = malloc(b->size());
b->clone(clonemem);
}
/Peter HTH,
Michiel Salters

[François Fleuret]

Dear John,

You wrote on 13 Mar 2006 12:51:47 MET:

This seems to work

#include <iostream>
using namespace std;

struct Base
{
int b;
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};
struct Derived : Base
{
int array[4];
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};

int main(int argc, char* argv[])
{
Base b;
Derived d;

Base * pb1 = &b;
Base * pb2 = &d;

pb1->Print_Size();
pb2->Print_Size();
return 0;
}

Yeah, but since sizeof used the static type of the class, you have to
re-write the method Print_Size -- with the exact same code -- in the
derived class.

I was wondering if there was a way to access the information about the
class size from its virtual table.

Cheers,

--
François Fleuret http://cvlab.epfl.ch/~fleuret

[Rolf Magnus]

François Fleuret wrote:

Dear John,

You wrote on 13 Mar 2006 12:51:47 MET:

This seems to work

#include <iostream>
using namespace std;

struct Base
{
int b;
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};
struct Derived : Base
{
int array[4];
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};

int main(int argc, char* argv[])
{
Base b;
Derived d;

Base * pb1 = &b;
Base * pb2 = &d;

pb1->Print_Size();
pb2->Print_Size();
return 0;
}
Yeah, but since sizeof used the static type of the class, you have to
re-write the method Print_Size -- with the exact same code -- in the
derived class.

That's right. You could use templates to avoid that:

template<typename T>
struct Base
{
int b;
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : Base<Derived>
{
//...
};
I was wondering if there was a way to access the information about the
class size from its virtual table.

No. The C++ standard doesn't even define the term "virtual table" or
anything similar.

[John Carson]

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:dv*************@news.t-online.com

François Fleuret wrote:

Dear John,

You wrote on 13 Mar 2006 12:51:47 MET:

This seems to work

#include <iostream>
using namespace std;

struct Base
{
int b;
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};
struct Derived : Base
{
int array[4];
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};

int main(int argc, char* argv[])
{
Base b;
Derived d;

Base * pb1 = &b;
Base * pb2 = &d;

pb1->Print_Size();
pb2->Print_Size();
return 0;
}

Yeah, but since sizeof used the static type of the class, you have to
re-write the method Print_Size -- with the exact same code -- in the
derived class.

That's right. You could use templates to avoid that:

template<typename T>
struct Base
{
int b;
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : Base<Derived>
{
//...
};

This works nicely for Derived objects. But how do you declare a Base object?

--
John Carson

[Rolf Magnus]

John Carson wrote:

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:dv*************@news.t-online.com

François Fleuret wrote:

Dear John,

You wrote on 13 Mar 2006 12:51:47 MET:

This seems to work

#include <iostream>
using namespace std;

struct Base
{
int b;
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};
struct Derived : Base
{
int array[4];
virtual void Print_Size()
{
cout << sizeof(*this) << endl;
}
};

int main(int argc, char* argv[])
{
Base b;
Derived d;

Base * pb1 = &b;
Base * pb2 = &d;

pb1->Print_Size();
pb2->Print_Size();
return 0;
}

Yeah, but since sizeof used the static type of the class, you have to
re-write the method Print_Size -- with the exact same code -- in the
derived class.

That's right. You could use templates to avoid that:

template<typename T>
struct Base
{
int b;
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : Base<Derived>
{
//...
};

This works nicely for Derived objects. But how do you declare a Base
object?

Oh, I forgot. You need a third class that is the base to all Base<> template
instances. Here is a complete and tested example:

#include <iostream>

using std::cout;
using std::endl;

struct Base
{
virtual void Print_Size() const = 0;
virtual ~Base() {}
};

template<typename T>
struct TBase : Base
{
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : TBase<Derived>
{
//...
int arr[32];
};

int main()
{
Base* test = new Derived;
test->Print_Size();
delete test;
}

[John Carson]

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:dv*************@news.t-online.com

John Carson wrote:

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:dv*************@news.t-online.com

That's right. You could use templates to avoid that:

template<typename T>
struct Base
{
int b;
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : Base<Derived>
{
//...
};

This works nicely for Derived objects. But how do you declare a Base
object?

Oh, I forgot. You need a third class that is the base to all Base<>
template instances. Here is a complete and tested example:

#include <iostream>

using std::cout;
using std::endl;

struct Base
{
virtual void Print_Size() const = 0;
virtual ~Base() {}
};

template<typename T>
struct TBase : Base
{
virtual void Print_Size() const
{
cout << sizeof(T) << endl;
}
};

struct Derived : TBase<Derived>
{
//...
int arr[32];
};

int main()
{
Base* test = new Derived;
test->Print_Size();
delete test;
}

You still haven't used a Base object. I am not sure in what way this version
improves on the earlier one.

Further, what happens with your scheme if you need a second level of
derivation: a Derived2 that derives from Derived?

--
John Carson