Understanding Inheritance in C++

Question

I am trying to learn C++, and wrote this code. According to my understaing this code need to produce output as "Derived Class" but output is "Base Class". Please help me understand this.

#include <iostream> 
using namespace std; 

class Base { 
    public: 
    char* name; 
    void display() { 
         cout << name << endl; 
    }

};

class Derived: public Base { 
   public: 
   char* name; 
   void display() { 
       cout << name << ", " << Base::name << endl; 
   } 
}; 

int main() { 
   Derived d; 
   d.name = "Derived Class"; 
   d.Base::name = "Base Class"; 

   Derived* dptr = &d; 

   Base* bptr = dptr; 

   bptr->display();
}

Please consider me as a beginner and explain why its output is "Base Class"

score 2 · Accepted Answer · answered Apr 04 '14 at 06:45

2

You need to make the display() method virtual

Like this:

class Base{ 
    public: 
    char* name; 
    virtual void display() { 
         cout << name << endl; 
 }

virtual allows derived classes to 'override' the base class' functions.

answered Apr 04 '14 at 06:45

AndyG

39,700
8
109
143

@andy.. After making it virtual its output is "Derived Class Base Class"...but I am expecting only "Derived Class" – someone Apr 04 '14 at 06:48
2

@someone: And why shouldn't it? You wrote `cout << name << ", " << Base::name << endl; ` not `cout << name << endl` – AndyG Apr 04 '14 at 06:51

score 0 · Answer 2 · answered Apr 04 '14 at 06:46

http://www.parashift.com/c++-faq/dyn-binding.html

Non-virtual member functions are resolved statically. That is, the member function is selected statically (at compile-time) based on the type of the pointer (or reference) to the object.

In contrast, virtual member functions are resolved dynamically (at run-time). That is, the member function is selected dynamically (at run-time) based on the type of the object, not the type of the pointer/reference to that object.

score 0 · Answer 3 · answered Apr 04 '14 at 06:47

0

If you want a derived class function to be called through a base class pointer, you need to make the base class function a virtual function. Add the key word virtual to the function and your code will be good to go.

answered Apr 04 '14 at 06:47

R Sahu

204,454
14
159
270

score 0 · Answer 4 · edited May 23 '17 at 12:07

0

C++ dispatch is little bit weird. Unless you declare the display method 'virtual' the bptr->display would alway call the display of base class.

More detailed explanation is here

edited May 23 '17 at 12:07

Community

1
1

answered Apr 04 '14 at 06:49

Vlad

9,180
5
48
67

score 0 · Answer 5 · answered Apr 04 '14 at 07:00

Polymorphism (working generically with base class pointers pointing to instances of derived classes) is done by dispatching to virtual functions as explained in the other answers. However, if you point to a derived class with a base class pointer, you will use the data members of the base class because the compiler doesn't know the data members of the derived class and same goes for calling functions of the base class, so the base class display() function uses the base class name data member and you result is "Base Class".

Understanding Inheritance in C++

5 Answers5