Hiding member function of Derived class from object of Derived class...

Hello,

Code:

---

class Base
{
  public:
      void myFun(){ "I m Base" };
};

class Derived : public Base
{
  public:
        new void myFun(){ "I m Derived" };     
        // This declaration is possible in C#.. and the similar thing i want to implement in C++.
};

int main()
{
    Base b;
    Derived d;
 
    b.myFun();
    d.myFun();      <<--
 
    return 0;
}

---

-------------------------------------------------------------------------------------------------------------------
Output:

I m Base
I m Base

Here, you can see the Derived::myFun() is hidden from object of Derived class i.e, d...
because i have declared myFun()....

Code:

---

new void myFun(){ .... }

---

<<--

So, how can i do this similar thing in C++..

Thanks & Regards
Kunal Nandi.

myFun() is non-virtual, so declaring a member function of the same name in Derived will hide myFun() from the base class. So you could declare a private myFun() member function in the Derived class, and the name hiding will take effect.

However, this will break the Liskov substitution principle. Why do you want to do this?

Currently i am Porting C# code to C++ (Qt).....
In that C# code there is one declaration of above type.. and i want to do the same in C++...
the declaration of code is as follows....

Code:

---

new public void SetCustomProperties(Hashtable properties)
{
    .........
    ......
    ..
}

---

so, the conclusion is tht i can't change the declaration of Derived Member function from public to private.. because it is somewhere called outside the class..

Please go through this link u will get wht excetly i want to ask.

http://msdn2.microsoft.com/en-us/lib...w2(VS.80).aspx

Quote:

---

so, the conclusion is tht i can't change the declaration of Derived Member function from public to private.. because it is somewhere called outside the class..

---

In that case just declare it public. The name hiding mechanism will still work.

you want me to make both member funtions of Base an Derived class Virtual.. is it so..???

Quote:

---

you want me to make both member funtions of Base an Derived class Virtual.. is it so..???

---

That depends on your design. I am just saying that name hiding will happen if you declare a derived class member function with the same name as a base class non-virtual member function. If you use virtual functions, name hiding will not happen, but the derived class member function will override the base class member function.

can you modify that code.. which i hv shown above.. according to ur view..

I m not gettting.. so please if possible then give any example code.. of name hiding.. or jst modify the code which i hv written..

Thanks..

Code:

---

class Base
{
  public:
      void myFun(){ std::cout << "I m Base" };
};

class Derived : public Base
{
  public:
        void myFun(){ std::cout << "I m Derived"; };     
};

int main()
{
    Base b;
    Derived d;
 
    b.myFun();
    d.myFun();
 
    return 0;
}

---

The above code will work [I fixed the myFun so that it's actually doing something].

It would be nicer to use a virtual function, because that way you can use a pointer for the base-object, and get access to the version of the object that you actually used for the instance, e.g:

Code:

---

class Base
{
  public:
      virtual void myFun(){ std::cout << "I m Base" };
};

class Derived : public Base
{
  public:
        virtual void myFun(){ std::cout << "I m Derived"; };     
};

int main()
{
    Base *b1;
    Base *b2;
    Derived d;
 
    b1 = new Base;
    b2 = new Derived;
    b1->myFun(); // Base::myFun()
    b2->myFun(); // Derived::myFun()
    d.myFun();    // Derived::myFun();
 
    return 0;
}

---

--
Mats