Claudio Roitman

Thanks Vanessa!

But it used to be there... didn't it?
What happened?
Claudio

Congrats!
You can ask for Your Logo at:
http://logos.sun.com/logosite.jsp?Category=third&Logo=cert-jtech&Page=click

Last week, I could see my SCJP score in the Sun certManager Web site. But Now, there is a new version of that site, and I can't find the score...
Have all scores gone? Or it just happened to my score?

Hi,
What is the natural next step after SCJP 1.4?
SUN CERTIFIED WEB COMPONENT DEVELOPER ?
SUN CERTIFIED BUSINESS COMPONENT DEVELOPER ?
or SUN CERTIFIED DEVELOPER FOR THE JAVA 2 PLATFORM?
Thanks,
Claudio

Thanks.
My advidse: Study a lot about Threads!!!

Hello,
And You can add something like this:
http://www.specialist.ru/trainers/logo/sun_java_programmer.gif

Hello!,
I passed SCJP 1.4 ! (Score 93%).
How did I study?:
-I took Sun SL-275 Course.
-I took The Sun Web Based free Tutorial
( http://java.sun.com/docs/books/tutorial/ )
-I read part of RHE Book
-Whizlabs Mock Tests.
- I use Eclipse Scrapbook Pages to test code snippets and also Eclipse to browse System sources
-I have not experience as a Java Programmer, but I have been an Smalltalk developer for 8 years, so I have a great Object Oriented background (I am also IBM Certified VisualAge Smalltalk developer).
- I found answers to ALL my questions and doubts in JavaRanch Forums archive
Thanks a lot to:
- JavaRanch
- Whizlabs
- Fer
- and Cristina
Claudio

If You want to be an Official Sun Teacher, Certification is a prerequisite!

Hi, Mock Exam practice number 9 from Whizlabs 4.0, Question number 27 says:
"Which of the following statements are true?
a) A Thread's start() method, automatically call the run() method
b)...
c)...
d)...
"
It says option a is true, But I think it is not, because start() method does not call run() method, It just put the receiver thread in a runnable state, and then, The Thread scheduler will decide when to invoke it's run() method...(maybe never!)
Am I rigth?
Thanks,
Claudio

I just wanted to show the difference.... I guess the explanation is:

Methods access variables only in context of the class of the object they belong to

.
It says: "access variables" but it is different for accessing methods.
But the afirmation:

(When an object is created, instances of all its super-classes are also created).

suggest the instance that actually executes m(), is an Instance of P (The superclass) but if it were an instance of P it would call P implementation of b() instead of Q implementation...

Muy Bien Mario!!!
Yo rindo Ma�ana y estoy muy NERVIOSO!!!

Originally posted by Mario Elkin Rodr�guez Alarc�n:

And don�t forget the ";" in anonymous classes .... ...

What do You mean?
BTW: Congratulations!

Originally posted by Claire Yang:
class P { int b=3; void m() { System.out.print(b); } } class Q extends P { int b=5; public static void main(String[] args) { Q q=new Q(); q.m(); System.out.println(", "+q.b); } } //output: 3, 5
In the above program, class Q's instance variable b hide class P's b and class Q inherits m() method from class P, how to explain that when object q calls m() method, it doesn't "see" its own b but class P's b?

...But if We have:
class P {
int b() {
return 3;
}
//instead of int b=3
void m() {
System.out.print(b());//instead of print(b)
}
}
class Q extends P {
int b() {
return 5;
}
//instead of int b=5
public static void main(String[] args) {
Q q=new Q();
q.m();
System.out.println(", "+q.b());
//instead of System.out.printl(", "+q.b); }
}
The output is 5,5 !!!

Yes Dennis, I was referring to Smalltalk.