Inheritance and Combination - Base class has a pointer to derived class - visual-c++

Question

I'm wondering if I can have two classes look like these:
//file: Small.h
#pragma once
#include "Little.h"
class Small :
public Little
{
public:
Small(void){}
~Small(void){}
};
and
//file: Little.h
#pragma once
#include <iostream>
#include "Small.h"
using namespace std;
//class Small;
class Little
{
public:
Little(){ s = 0; }
void print(){ cout << "oops!" << endl; }
Small* s;
};
And now my problem: When I wanna create an object of type "Small" and call its "print()" function, VS-2010 says that "class 'Small' has no member named 'print()'." What's the solution?

Answer 1

The following should work, however, you should be constructing classes in this manner, the base class should really have a member of the type of a derived class.
However as the member is just a pointer, the code should still compile and work, however your asking for errors.
The problem with you code is that the class Small must be defined before you create the class Little and Little bfroe you create Small
You Should do it in one file as follows, as they are interdependent
//file: SmallLittle.h
#pragma once
#include <iostream>
using namespace std;
class Small;
class Little
{
public:
Little(){ s = 0; }
void print(){ cout << "oops!" << endl; }
Small* s;
};
class Small :
public Little
{
public:
Small(void){}
~Small(void){}
};
However since Small is a member type of Little, your probably better of creating a single class as follows, The only reason for using your code is so you can use Little code in which Small is not defined, or used. However in this case your better off using a void * type for the pointer.
as Follows
//file: Small.h
#pragma once
#include <iostream>
using namespace std;
class Small;
class Small
{
public:
Small(){ s = 0; }
void print(){ cout << "oops!" << endl; }
Small* s;
~Small(void){}
};
Also you I don't think you need to set s to 0, as this is NULL, which should be the value until you create a pointer using new of assign a pointer.