About Question enthuware.ocajp.i.v7.2.907 :

[mistervernon]

Hi (awesome product!).
...your explanation to this question reads:
"Note that if a subclass class constructor doesn't explicitly call the super class constructor, the compiler automatically inserts super(); as the first statement of the base class constructor. So option 5 is not needed."

However, I believe that the explanation should read:
"Note that if a subclass class constructor doesn't explicitly call the super class constructor, the compiler automatically inserts super(); as the first statement of the subclass class constructor. So option 5 is not needed."



.... I think!?

[admin]

Hi,
You don't seem to have the latest version of the question bank because I see that it has already been fixed.

HTH,
Paul.

[JaredTse]

I just need a minor clarification on this. My understanding of inheritance is that If a Subleases does not have any constructors, then the compiler inserts a default no args constructor which calls the Supperclass no args Constructor. However if you supply a constructor in this case a constructor with a single arg. Then the compiler does not insert the default constructor.

Then my question is, how is the superclass constructor been called ?
is there super() method inserted at compile time on the Constructor that takes a single arg on subclass?
and also does the super() method gets inserted on all the constructors at compile time regardless of number of args ?
Finally I do not see the super() method on the .class files, is it supposed be there ?

Thanks

[admin]

Yes, if you don't explicitly call any superclass constructor in your constructor, then the compiler automatically inserts a call to super(); at the first line.
super(); is not a method and is not a method call. It is a syntax to call the super class's constructor that takes no arguments.

HTH,
Paul.

[daromnik]

Hi.
But this code doesn't compile, because "No "public class" found to execute".
Isn't that right?

[admin]

1. A class doesn't need to be public for successful compilation.
2. "No "public class" found to execute" means you are trying to execute it. Execution is after compilation. That means, it has already compiled.
3. A class doesn't need to be public for execution either. Only the main method needs to be public.
4. You seem to be using an IDE. Please use compile and run it from command line. IDEs are known to cause confusion by showing custom error messages and are not recommended while preparing for the exam.

[daromnik]

Thank you very much!

[en.builin@gmail.com]

Make Eagle class declaration public:
public class Eagle { ... }

It can be right answer. Class declaration doesn't have extends keyword, so super() wouldn't be called. Filename is Eagle.java, so it will compile.
Am I right?

[admin]

You are right. The option is missing the extends part and without the extends part, it would make the code valid. Should be fixed.
thank you for your feedback!