Initialize an object with a different class

Question

I saw this example: http://www.cplusplus.com/doc/tutorial/typecasting/#dynamic_cast

(...)
class Base { virtual void dummy() {} };
class Derived: public Base { int a; };
(...)
Base * pba = new Derived;
Base * pbb = new Base;
(...)

Why 'pba' is a Base object if it's being initialized with Derived? Why not make it a Derived object?

Derived * pba = new Derived; // use this instead

And is it just a C++ thing?

That's an instance of *polymorphism*. You cannot expect us to provide a fully elaborate answer as this is a pretty broad subject (and I'm sure there are some duplicates here). — cadaniluk, Jan 09 '17 at 12:51
And it's an example for dynamic_cast. If you write _Derived * pba = new Derived;_ then this line has no sense _pd = dynamic_cast(pba);_ — rud0x1, Jan 09 '17 at 12:53
If you have both [cats and dogs in a set](https://en.wikipedia.org/wiki/Polymorphism_%28computer_science%29%23Subtyping), and want to make all animals in this set make a sound then you call `Animal::talk()` for each animal, but this call will yield different results, based on the actual class of the animal. — Dialecticus, Jan 09 '17 at 12:57
`pba` is not a `Base` object, it's a *pointer* to a `Base` object. It is what it is in order to illustrate the subject of `dynamic_cast`. It seems like you have looked at the code without reading the surrounding text. — molbdnilo, Jan 09 '17 at 13:00
The other thing is that this is a simplified example. A (slightly) more realistic version would be: `Base*pba = flag ? new Derived: new Derived2;` (where `Derived2` is a different derived class). — Martin Bonner supports Monica, Jan 09 '17 at 13:18
@molbdnilo, yes I did, but i'm not interested on the dynamic_cast, but in what situation declaring a object with a class and initializing with another would be used, and Dialecticus answered one. The link was just and example — Lucas Steffen, Jan 09 '17 at 13:24
@Downvoter the question is Why not make pbb a Derived object? — Lucas Steffen, Jan 09 '17 at 13:28
Because you're referencing an *example*, which demonstrates a feature of the language. Making `pbb` a `Derived` object would defeat the purpose of the example. If you want a real-life example for polymorphism, google it. It's an inherent feature of OOP and is explaines in virtually every OOP tutorial. [Here](http://stackoverflow.com/questions/1031273/what-is-polymorphism-what-is-it-for-and-how-is-it-used) is a question on polymorphism with extensive answers. — cadaniluk, Jan 09 '17 at 13:32

score 1 · Accepted Answer · answered Jan 09 '17 at 13:46

neither pba nor pbb is an object but they are pointers of type base class Base so in your code you used he pointers polymorphically which means a base pointer can point to the same class or to its derived class object.

the object is created with new not pbb or pba themselves, consider this example:

#include <string>
#include <iostream>
using namespace std;


class Base
{
    public:
        virtual void Print() const { cout << "Base print()" << endl;} // virtual function
        void Greet()const {cout << "in Base Say: hello!" << endl;}
};

class Derived : public Base
{
    public:
        void Print() const { cout << "Derived print()" << endl;} // ovrode the base member pritn()
        void Greet()const {cout << "in Derived Say: hello!" << endl;}
};

int main()
{

    Base* pba = new Derived;

    pba->Print(); // Derived print()
    pba->Greet(); // in Base Say: hello! ??? because Greet() is not virtual

    Base* pbb = new Base;

    pbb->Print(); // Base print()
    pbb->Greet(); // in Base Say: hello!

    return 0;
}

so at runtime the pointer pba and pbb can be assigned an object of Base or Derived classes thus the virtual member function are called accordingly.

Initialize an object with a different class

1 Answers1