"new" modifier explicitly hide a member inherited from a base class in C#

[kunalnandi]

Hello,

"new" modifier explicitly hide a member inherited from a base class in C#..
Is there any alternative in C++ which can be used like this..
as shown below..

public class MyBaseC 
{
   public int x;
   public void Invoke() {}
}

public class MyBaseC 
{
   public int x;
   public void Invoke() {}
}

To copy to clipboard, switch view to plain text mode 

Declaring a member with the name
Invoke
in a derived class will hide the method
Invoke
in the base class, that is:

public class MyDerivedC : MyBaseC
{
   new public void Invoke() {}
}

public class MyDerivedC : MyBaseC
{
   new public void Invoke() {}
}

To copy to clipboard, switch view to plain text mode 

here member function Invoke() in class MyDerivedC is hidden from the base MyBaseC...

-->> how can i implement this similar thing in C++...

Regards,
Kunal nandi.

[wysota]

There are many ways. The simplest is to prefix all member variables, for example "m_invoke" instead of "invoke". Another is to use a separate internal namespace:

class x {
public:
//...
  void invoke() { xpriv::invoke++; }
private:
  namespace xpriv {
    int invoke;
  }
};

class x {
public:
//...
  void invoke() { xpriv::invoke++; }
private:
  namespace xpriv {
    int invoke;
  }
};

To copy to clipboard, switch view to plain text mode 

You can also use a data-pointer and hide all member variables in a separate object.

[jacek]

here member function Invoke() in class MyDerivedC is hidden from the base MyBaseC...

-->> how can i implement this similar thing in C++...

What do you mean exactly by "hidden from the base MyBaseC"? But it seems that you just want the default behaviour of non-virtual methods:

#include <iostream>
 
class Base
{
   public:
      void foo() { std::cout<< "Base::foo" << std::endl; }
      virtual void bar() { std::cout<< "Base::bar" << std::endl; }
};
 
class Derived : public Base
{
   public:
      void foo() { std::cout<< "Derived::foo" << std::endl; }
      virtual void bar() { std::cout<< "Derived::bar" << std::endl; }
};
 
int main()
{
   Derived d;
   Base *b = &d;
 
   d.foo();
   b->foo(); // <--
 
   d.bar();
   b->bar();
 
   return 0;
}

#include <iostream>

class Base
{
   public:
      void foo() { std::cout<< "Base::foo" << std::endl; }
      virtual void bar() { std::cout<< "Base::bar" << std::endl; }
};

class Derived : public Base
{
   public:
      void foo() { std::cout<< "Derived::foo" << std::endl; }
      virtual void bar() { std::cout<< "Derived::bar" << std::endl; }
};

int main()
{
   Derived d;
   Base *b = &d;

   d.foo();
   b->foo(); // <--

   d.bar();
   b->bar();

   return 0;
}

To copy to clipboard, switch view to plain text mode 

Output:

Derived::foo
Base::foo
Derived::bar
Derived::bar

[kunalnandi]

What do you mean exactly by "hidden from the base MyBaseC"? But it seems that you just want the default behaviour of non-virtual methods:

#include <iostream>
 
class Base
{
   public:
      void foo() { std::cout<< "Base::foo" << std::endl; }
      virtual void bar() { std::cout<< "Base::bar" << std::endl; }
};
 
class Derived : public Base
{
   public:
      void foo() { std::cout<< "Derived::foo" << std::endl; }
      virtual void bar() { std::cout<< "Derived::bar" << std::endl; }
};
 
int main()
{
   Derived d;
   Base *b = &d;
 
   d.foo();
   b->foo(); // <--
 
   d.bar();
   b->bar();
 
   return 0;
}

#include <iostream>

class Base
{
   public:
      void foo() { std::cout<< "Base::foo" << std::endl; }
      virtual void bar() { std::cout<< "Base::bar" << std::endl; }
};

class Derived : public Base
{
   public:
      void foo() { std::cout<< "Derived::foo" << std::endl; }
      virtual void bar() { std::cout<< "Derived::bar" << std::endl; }
};

int main()
{
   Derived d;
   Base *b = &d;

   d.foo();
   b->foo(); // <--

   d.bar();
   b->bar();

   return 0;
}

To copy to clipboard, switch view to plain text mode 

Output:

Hello,

d->bar();

this will call the member function of Derived class...
but, wht is want is.. member function of Derived class Derived::bar() should be hidden from the derived class object.... and the object of Derived class should called Base::bar() member function...

this is possible in C# if we declare the derived class member function like as follow...


// In C#
class Base
{
public:
void foo() { std::cout<< "Base::foo" << std::endl; }
virtual void bar() { std::cout<< "Base::bar" << std::endl; }
};

class Derived : public Base
{
public:
void foo() { std::cout<< "Derived::foo" << std::endl; }
new void bar() { std::cout<< "Derived::bar" << std::endl; } <---
};


int main()
{
Derived d;

d.bar(); <---

return 0;
}

Output:

Base::bar();

[jacek]

member function of Derived class Derived::bar() should be hidden from the derived class object.... and the object of Derived class should called Base::bar() member function...

Are you talking about this?

#include <iostream>
 
class Base
{
   public:
      void foo() { std::cout << "Base::foo" << std::endl; }
      void bar() { std::cout << "Base::bar" << std::endl; }
};
 
class Derived : public Base
{
   public:
      void foo() { std::cout << "Derived::foo" << std::endl; }
};
 
int main()
{
   Derived d;
 
   d.foo();
   d.bar(); // <--
 
   return 0;
}

#include <iostream>
 
class Base
{
   public:
      void foo() { std::cout << "Base::foo" << std::endl; }
      void bar() { std::cout << "Base::bar" << std::endl; }
};
 
class Derived : public Base
{
   public:
      void foo() { std::cout << "Derived::foo" << std::endl; }
};
 
int main()
{
   Derived d;

   d.foo();
   d.bar(); // <--
 
   return 0;
}

To copy to clipboard, switch view to plain text mode 

Output:

Derived::foo()
Base::bar()