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How does default virtual behavior differ in C++ and Java ?

Default virtual behavior of methods is opposite in C++ and Java:

In C++, class member methods are non-virtual by default. They can be made virtual by using virtual keyword. For example, Base::show() is non-virtual in following program and program prints “Base::show() called”.

#include<iostream>
  
using namespace std;
  
class Base {
public:      
  
    // non-virtual by default
    void show() {  
         cout<<"Base::show() called";
    }
};
  
class Derived: public Base {
public:      
    void show() {
         cout<<"Derived::show() called";
    }      
};
  
int main()
{
  Derived d;
  Base &b = d;   
  b.show(); 
  getchar();
  return 0;
}

Adding virtual before definition of Base::show() makes program print “Derived::show() called”



In Java, methods are virtual by default and can be made non-virtual by using final keyword. For example, in the following java program, show() is by default virtual and the program prints “Derived::show() called”

class Base {
  
    // virtual by default
    public void show() {
       System.out.println("Base::show() called");
    }
}
  
class Derived extends Base {
    public void show() {
       System.out.println("Derived::show() called");
    }
}
  
public class Main {
    public static void main(String[] args) {
        Base b = new Derived();;
        b.show();
    }
}

Unlike C++ non-virtual behavior, if we add final before definition of show() in Base , then the above program fails in compilation.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

tags:

Java Java

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