Re: acces of members of subclass

Lew wrote:
The call to super() is *always* present.  Any given class, in the
general case, can rely on the presence of super() call.  Daniel and
you are not correct.

Eric Sosman <Eric.Sos...@xxxxxxx> wrote:
     No, Mark is right, and Daniel and Lew are wrong.

I said that there was a call. I never claimed that it would always
succeed.

     Lew (and the JLS) are right in saying that the compiler inserts a
call to super() if the constructor does not begin with a call to one

How can I be wrong if I said the same thing that the JLS does?

of its superclass' constructors.  But that does not guarantee that the
superclass actually *has* a no-arguments constructor that can be called
as plain super().  If it doesn't, there'll be a compile-time error.

I never said otherwise.

Try it!

The error that results proves that I was correct. There cannot be an
error on a call to super() unless there's a call to super() to have an
error.

--
Lew



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Relevant Pages

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