Dynamic Object Not Creating for Privately Inherited Class.

Question

Hi,

What is the reason for the following code that does not let me to create object.

class base
{
    public:
        void foo()
        { cout << "base::foo()"; }
};

class derived : private base
{
    public:
        void foo()
        { cout << "deived::foo()"; }
};

void main()
{
    base *d = new derived();
    d->foo();
}

It Gives me error :

" 'type cast' : conversion from 'derived *' to 'base *' exists, but is inaccessible"

Thanks in advance :)

Answer 1

The problem is that you are using private inheritance; this means that inheritance can only be seen inside your class (in this case, derived). You cannot point a base* to a derived instance outside your class (in this case, in main()) because the inheritance (and hence, the conversion) cannot be accessed.

This is exactly the same as trying to access a private member from outside a class.

In fact, the name "private inheritance" is quite misleading, since it does not implement real inheritance. In your example, a derived instance is not a base; it is just implemented in terms of a base, and this is what "private inheritance" means. If you are tempted to use private inheritance, you should consider the possibility of using simple aggregation (i.e.: a private pointer to base inside derived) instead. In most cases (most cases, not always), private inheritance offers no advantages and has some subtle problems.

Answer 2

Private inheritance causes your base class details to not to visible outside of derived class. so the error C2243: Access protection (protected or private) prevented conversion from a pointer to a derived class to a pointer to the base class.

You should use public inheritance instead.

Answer 3

Two ways to make it work:

class base
{
    public:
        void foo()
        { std::cout << "base::foo()"; }
};

class derived : public base // <-- make it public
{
    public:
        void foo()
        { std::cout << "derived::foo()"; }
};

void main()
{
    base *d = new derived();
    d->foo();
}

or:

class base
{
    public:
        void foo()
        { std::cout << "base::foo()"; }
};

class derived : private base
{
    public:
        void foo()
        { std::cout << "derived::foo()"; }
};

void main()
{
    derived *d = new derived(); // create pointer on derived, not base
    d->foo();
}

Hope that helps, Exa

Answer 4

My guess: the default constructor/copy constructor of your base class created by your compiler is private due to private inheritance to your derived class and can't by used by the type cast

Answer 5

Because you used private inheritance, a derived object is not a kind of base object. So when you say:

 base *d = new derived();

the pointer to a derived cannot be converted to a pointer to a base - there is no real relationship between the two. You can create derived objects though:

derived d;
derived * dp = new derived();