Question

Suppose I have the following hierarchy:

class A  
{  
 public:  
 A()   
 private:  
 int aa;  
}

class B: public A  
{  
 public:  
 B()   
 private:  
 int bb;  
}

class D: public B  
{  
 public:  
 D()   
 private:  
 int dd;  
}

When I type in the following code in my main:

D dobj;
std::cout<<"Address of D object: "<<&dobj<<std::endl;

A aobj = static_cast<A>(dobj);
A* aptr = static_cast<A*>(&dobj);

std::cout<<"Address of D object: "<<&dobj<<std::endl;
std::cout<<"Address of aptr object: "<<&aptr<<std::endl;
std::cout<<"Address of A object: "<<&aobj<<std::endl;

The output of which is:

Address of dobj object: 0012FF0C
Address of dobj object: 0012FF0C
Address of aptr object: 0012FF18
Address of aobj object: 0012FF14

Why is the address of aptr and aobj different ?? Aren’t they supposed to be the same ??

And why is the address of dobj and aptr different ?? Aren’t they supposed to be same as well ??

I am compiling on windows using VC++.

Thanks,
De Costo.

Answer 1

Why are aptr and &aobj different?

First, you are printing out the wrong pointer: &aptr is “the address of aptr,” not “the address of the object to which aptr points.” You need to print just aptr to print “the address of the object to which aptr points,” which is presumably what you really want to do.

aptr is a pointer to the A part of dobj. aobj is a copy of the A part of dobj.

aptr and &aobj are different because they are the addresses of different objects.