CODEWITHSUNDEEP
c++
user3759947
user3759947

Reputation: 13

How to access the function which is defined in derived class and not in base class in polymorphism

How to access the function which is defined in derived class and not in base class in polymorphism?

class base {
    public:
        virtual void display ()
        {
            cout << "base" << endl;
        }
};

class derived : public base {
    public:
        void display (){
            cout << "Derived" << endl;
        }

        void goo (){
            cout << " new function in derived" << endl;
        }
};

base * global_function ( base *ptr)
{
    /**** how to invoke the derived class goo function  ---Here *****/


}

int main ()
{

    derived obj;
    global_function ( &obj );

}

Can anyone help me how to invoke the derived class function, which is not specified in the base class?

Upvotes: 1

Views: 75

Answers (4)

MABVT
MABVT

Reputation: 1360

The solution ( dirty and not recommended ) was the following

base* global_function ( base *ptr ) {
    std::dynamic_cast<Derived>(*ptr).goo();
}

Although this might work in this specific case, just don't assume a base type pointer to contain information about foo, rather do template specialization and maintain typesafety!

template <typename T>
T* global_function(const T* ptr) {
    // do nothing in this case. 
}
// Specialization
template<>
base* global_function<base>(const base* ptr) {
    // invoke base methods
}

template<>
Derived* global_function<Derived>(const Derived* ptr) {
    ptr->goo();
}

Upvotes: 0

Bathsheba
Bathsheba

Reputation: 234695

You can access the method but you have to hope that base *ptr is actually a derived* type (which, in your case, it is).

Then you can use a dynamic_cast to convert the pointer to the correct type:

derived* my_ptr = dynamic_cast<derived*>(ptr)

and call the method using my_ptr->goo(). If my_ptr evaluates to nullptr then ptr was not of the correct type. You'll need to test this else before calling goo else you could invoke undefined behaviour.

But, this is an anti-pattern that circumvents polymorphism. It's far better to set up your class structure so you can avoid this cast.

Upvotes: 1

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726569

Like this:

Derived *d = dynamic_cast<Derived*>(ptr);
if (d) d->goo();

dynamic_cast<Derived*> will produce a valid Derived* pointer if ptr indeed points to Derived; otherwise, NULL pointer would be produced (note the if that "guards" the invocation of goo()).

In general, it is not a good idea to put dynamic_cast to excessive use: if you need to do it a lot, it points to a potential shortcoming in the design of your class hierarchy.

Upvotes: 2

Orelsanpls
Orelsanpls

Reputation: 23515

You can downcast the base pointer

Look at Here, this contain a very good explanation of what to do

Taked from the link:

Child *p = dynamic_cast<Child *>(pParent);

Upvotes: 2

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