Accessing base class's method with derived cla

Question

when accessing foo() of "base" using derived class's object.

#include <iostream>

class  base
{
      public:
      void foo()
     {
        std::cout<<"\nHello from foo\n";
     }
};

class derived : public base
{
     public:
     void foo(int k)
     {
        std::cout<<"\nHello from foo with value = "<<k<<"\n";
     }
};
int main()
{
      derived d;
      d.foo();//error:no matching for derived::foo()
      d.foo(10);

}

how to access base class method having a method of same name in derived class. the error generated has been shown. i apologize if i am not clear but i feel i have made myself clear as water. thanks in advance.

Answer 1

You could add using base::foo to your derived class:

class derived : public base
{
public:
     using base::foo;
     void foo(int k)
     {
        std::cout<<"\nHello from foo with value = "<<k<<"\n";
     }
};

Edit: The answer for this question explains why your base::foo() isn't directly usable from derived without the using declaration.

Answer 2

d.base::foo();

Answer 3

Add following statement to your derived class :

using Base::foo ;

This should serve the purpose.

When you have a function in derived class which has the same name as one of the functions in base class, all the functions of base class are hidden and you have to explicitly bring them in scope of your derived class as mentioned.

Answer 4

we can call base class method using derived class object. as per below code.

#include<iostream.h>
class base
{
public:
void fun()
{
    cout<<"base class\n";
}
};
class der : public base
{
public:
void fun()
{
cout<<"derived class\n";
}
};
int main()
{

der d1;


d1.base::fun();


return 0;
} 

Output will be base class.