omkar1010

 Profile Answers by omkar1010 

• Jun 25th, 2008

 

yes we can access it as public function

jezpops

 Profile Answers by jezpops 

• Jul 7th, 2008

 

I dont understand what you want to do by this question. Usually the virtual keyword would be used in the base class and the method overridden in the derived class.

The virtual keyword ensures that the correct overridden function in the derived class is called rather than the function in the base class. It ensures only one instance of the object exists at runtime.

Since it only usually used in inheritance then methods of the base class would be accessible anyway (dependent on the access specifiers)

arun_goel3

 Profile Answers by arun_goel3 

• Jul 15th, 2008

 

using typecasting and virtual we can access the private member of base class

#include <iostream>

using namespace std;

class base

{

int i;virtual void update(int j)

{

cout << "in base update" << endl;

i = j;

}

public :

base ( int ii = 10) : i(ii) {}

void print()

{ cout << "value of i " << i << endl; }

virtual ~base() {}

};

class derived : public base

{

public :

int i;derived (

int i1 = 20, int i2 = 21) : base(i1) , i(i2) {}virtual void update(int j)

{

cout << "in derived update" << endl;

i = j;

}

};

void main()

{

base *b1 = new base();

b1->print();

derived *d = (derived *)b1;

d->update(30);

b1->print();

delete b1;

}

burraganesh

 Profile Answers by burraganesh 

• Dec 7th, 2008

 

No way,
Ex:
#include<iostream>
using namespace std;
class base{
    private:
        int a;
    public:
        virtual void show()
        {
            cout<<a<<endl;
        }
};
class derived:public base{
    public:
        void display()
        {
            cout<<a<<endl;//we must call show() here instead
        }
};
int main()
{
    derived ob;
    ob.display();
    return 0;
}