- In Java, every method is like virtual method of C++,
- Constructors don't get called in nested heirarchy in Java. You need to
call the base class constructor explicitly
Your c++ code will be equivalent to java code if you make following changes:
Java Code :
class A {
A() { this.show(); }
void show() { System.out.println("Base Class"); }
};
class B extends A {
B() { super(); }
void show() { System.out.println("Derived Class"); }
};
public class this_test {
public static void main(String args[]) { System.out.println("This is
Main");
B test = new B(); }
};
C++ :
class A {
public:
A() { this->show(); }
virtual void show() { cout<<"Base Class"<<endl; } };
class B : public A {
public:
B() { this->show(); }
void show() { cout<<"Derived Class"<<endl; } };
int main(void) { cout<<"This is main"<<endl; B test; }
> Hi all,
>
> Today I came across a very interesting thing, It is really
amazing
> to me.
> This may sound very silly to you. But tell me why java behaving in this
> fashion. I feel the behaviour of C++ is correct.
>
>
> PS: I may or may not be in this groups, so please include my email-id
too
> in your reply.
>
> Java Code :
> class A {
> A() { this.show(); }
> void show() { System.out.println("Base Class"); }
> };
>
> class B extends A {
> B() { }
> void show() { System.out.println("Derived Class"); }
> };
>
> public class this_test {
> public static void main(String args[]) { System.out.println("This is
Main");
> B test = new B(); }
> };
>
> Output :
> This is Main
> Derived Class
>
> C++ code:
> class A {
> public:
> A() { this->show(); }
> virtual void show() { cout<<"Base Class"<<endl; }
> };
>
> class B : public A {
> public:
> B() { }
> void show() { cout<<"Derived Class"<<endl; } };
>
> int main(void) { cout<<"This is main"<<endl; B test; }
>
> Output:
> This is main
> Base Class
>
>
> Thanks,
> Balabaskaran.
>
>
>
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