why size of *this and size of basic and derived class different?

[Bob]

Try the following code
class basic
{
public:
virtual void one()=0;
};

class derived1:public basic
{
public:
virtual void two()=0;
};

class derived2:pubilc base
{
public:
virtual void three()=0;
};

class component : public derived,public derived2
{
public:
void one(){ cout<<sizeof(*t his);}
void two(){cout<<siz eof(*this);}
void three(){cout<<s izeof(*this);}


};

void main()
{
//
basic * b = (derived1*)new component;
cout<<*b;
b->one();

}

My question is why size of *b and size of *this are different.
sizeof(*b) will give the size of the base class (4)while
sizeof(*this)gi ves the size of component class(8) although the pointer
used here is *b i.e basic class pointer.

regards,
Bob

[John Harrison]

"Bob" <bo**********@r ediffmail.com> wrote in message
news:e8******** *************** ***@posting.goo gle.com...

Try the following code
class basic
{
public:
virtual void one()=0;
};

class derived1:public basic
{
public:
virtual void two()=0;
};

class derived2:pubilc base
{
public:
virtual void three()=0;
};

class component : public derived,public derived2
{
public:
void one(){ cout<<sizeof(*t his);}
void two(){cout<<siz eof(*this);}
void three(){cout<<s izeof(*this);}


};

void main()
{
//
basic * b = (derived1*)new component;
cout<<*b;
I guess you mean

cout<<sizeof(*b );
b->one();

}

My question is why size of *b and size of *this are different.
sizeof(*b) will give the size of the base class (4)while
sizeof(*this)gi ves the size of component class(8) although the pointer
used here is *b i.e basic class pointer.

That's not right, the method one() is in the Component class so 'this' has
type Component*, therefore sizeof(*this) equals sizeof(Componen t).

Not sure why you think it would be any different, but remember that sizeof
is worked out at compile time. The number printed by the method one() will
always be the same no matter what pointer is used to call it.

john