Casting changes value of ptr

[Vinodh Kumar]

I see that casting changes the value of a pointer in case of multiple
inheritance.In single inheritance also it is the same know?Isn't it?

Vinodh Kumar P

 

[John Harrison]

I see that casting changes the value of a pointer in case of multiple
inheritance.In single inheritance also it is the same know?Isn't it?

Vinodh Kumar P

It can happen, for instance if the derived class has virtual functions and
the base class does not.

#include <iostream>
using namespace std;

class A
{
int a;
};

class B : public A
{
virtual ~B() {}
int b;
};

int main()
{
B* b = new B;
A* a = b;
cout << a << ' ' << b<< '\n';
}

Basically this is an area where you shouldn't mind if a cast or conversion
changes the pointer value. Do you think its a problem, or are you just
trying to learn more about C++?

john

 

[Kevin Goodsell]

It can happen, for instance if the derived class has virtual functions and
the base class does not.

But having a base class with no virtual destructor is asking for trouble.

#include <iostream>
using namespace std;

class A
{
int a;
};

Note no virtual destructor.

class B : public A
{
virtual ~B() {}
int b;
};

int main()
{
B* b = new B;
A* a = b;
cout << a << ' ' << b<< '\n';

Assuming you want to prevent memory leaks, you'd obviously need to free
the allocated object. This:

delete a;

Creates undefined behavior. In this particular case it's easy to work
around. You just do this instead:

delete b;

But in general you may have a pointer of type A* which may point to some
derived class, which is not known. The only safe way to delete it (in
the absence of a virtual destructor) is to first cast it to the exact
correct pointer type. This can obviously be a major hassle.

}

Basically this is an area where you shouldn't mind if a cast or conversion
changes the pointer value. Do you think its a problem, or are you just
trying to learn more about C++?

I agree - usually, the exact value of a pointer should not matter.

-Kevin

 

[Vinodh Kumar]

----- Original Message -----
From: "Kevin Goodsell" <[email protected]>
Newsgroups: comp.lang.c++
Sent: Thursday, July 31, 2003 12:22 PM
Subject: Re: Casting changes value of ptr

But having a base class with no virtual destructor is asking for trouble.


Note no virtual destructor.


Assuming you want to prevent memory leaks, you'd obviously need to free
the allocated object. This:

delete a;

Creates undefined behavior. In this particular case it's easy to work
around. You just do this instead:

delete b;

But in general you may have a pointer of type A* which may point to some
derived class, which is not known. The only safe way to delete it (in
the absence of a virtual destructor) is to first cast it to the exact
correct pointer type. This can obviously be a major hassle.


I agree - usually, the exact value of a pointer should not matter.

But the type of the object that is being pointed to difers in case of a
casting in multiple inheritance.Thats the same case when we cast between
objects of single inheritance hiearchies.So I conclude that both in single
and multiple inheritance hierarchies, casting changes the value of the
pointer.I agree that the exact value of the point doesn't matter.

 

[Vinodh Kumar]

In single inheritence, in common implementations the cast shouldn't
change the value of the pointer. Imagine:

class A
{
int i;
virtual ~A();
};

class B: public A
{
int j;
};

B* b = new B;
A* a = b;


Now, the layout of a B might look like this:

vtable ptr (4 bytes)
A part: int i; (4 bytes)
B part: int j; (4 bytes)

Now, if the B object is created at the address 10, say, then the
object will run from address 10 though to 22.

The value of b above will be 10.
The value of a above will also be 10!

IOW, the value of the pointer will not change in single inheritence
cases.

class A
{
int i;
virtual ~A();
};

class B: public A
{
int j;
};

class C: public B
{
int k;
}

Now, the layout of a C might look like this:

vtable ptr (4 bytes)
A part: int i; (4 bytes)
B part: int j; (4 bytes)
C part: int k;(4 bytes)

Now, if the B object is created at the address 10, say, then the
object will run from address 10 though to 24.

C* c = new C;
A* a = C;

The value of c above will be 10.
The value of a above will also be 10.
Fine.

B* b = c;
Now the value of b will be different.
So in single inheritence casting changes the value of the pointer if we are
not casting to top most base class or bottom most derived class.

 

[tom_usenet]

class A
{
int i;
virtual ~A();
};

class B: public A
{
int j;
};

class C: public B
{
int k;
}

Now, the layout of a C might look like this:

vtable ptr (4 bytes)
A part: int i; (4 bytes)
B part: int j; (4 bytes)
C part: int k;(4 bytes)

Ok, there are certainly implementations that will use that layout.

Now, if the B object is created at the address 10, say, then the
object will run from address 10 though to 24.

?? You don't create a B object on the line below, but a C object.

C* c = new C;
A* a = C;

The value of c above will be 10.
Ok.

The value of a above will also be 10.
Fine.
Ok.


B* b = c;
Now the value of b will be different.

Why? It will still be 10 on normal implementations (I don't know of an
implementation where it will change, and there may well not be one,
although the standard allows it). The B object doesn't care whether
it's a subobject of a C object, a D object or whether it isn't a sub
object at all, but the whole object.

So in single inheritence casting changes the value of the pointer if we are
not casting to top most base class or bottom most derived class.

Why do you think that the value will change in the above case? It
wouldn't make sense for it to change. Try this on your favourite
compilers:

class A
{
public:
int i;
virtual ~A(){}
};

class B: public A
{
public:
int j;
};

class C: public B
{
public:
int k;
};

#include <iostream>

int main()
{
C* c = new C;
B* b = c;
A* a = c;

std::cout << (void*)a << '\n';
std::cout << (void*)b << '\n';
std::cout << (void*)c << '\n';
}

Tom