About SCJP!

Hey Bros,
I have a good programming background(C#.Net and VB.Net)
Frankly, I'm kindda bored of Microsoft stuff, So I'm Planning on SCJD.
Of Course, we all know that you should have a SCJP first, So I have two ways to accomplish that :
1-Self-Learning: depending On my previous experience in programming, and most at my ex in C#, I can go through Java and learn what makes me qualified for a SCJP exam or,
2-Training Center: I go to one of those, take the essential training of SCJP,
and spend a lot of money(Which is Something that I don't prefer!)
Plus that these kind of courses contain some materials that became basics to me ..
Here's the Java Diploma tracks from the training center I went:
-Java Programming Language basics
-Advanced Object Oriented Programming
-Building Java Applications
-Developing Graphical User Interfaces
-Advanced Java Programming
So The questions are :
-Which Way to Choose?
and If I Choose the self-learning way..
-When Can I tell that I'm Qualified for popping into a SCJP certification exam ??
Thanks In Advance!

There is no need for an overpriced training center to do the SCJP. What you need is to get a good text, look for something by by Kathy Sierra and Bert Bates that is appropriate for the cert you want to do. Do all the exercises and any Mock Exams in the book.
Go to www.javaranch.com (K&B's website) and another Java Forum. they have extensive coverage of all certification topics, with a forum per exam and lots of activity. they can recommend plenty of further mock tests.
You are ready when you can get a nearly perfect score on the mock tests.
And there is absolutely no need to pay money to some scumbag organization that claims a 150% pass rate to get mock tests.

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    Thanks -- I finally collated some good tips a few days after I posted that here, with some help from local LUGgers. I'll post them here for the archives:
    Would you mind posting a summary of the helpful tips, tricks, and
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    1. The unary bitwise NOT operator (~) has a shortcut, but one which
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    Several ways were suggested:
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    123 / 8 = 15 remainder 3 -- 3 is rightmost digit (3)
    15 / 8 = 1 remainder 7 -- 7 is next to 3 (73)
    we don't divide 1 by 8 -- 1 is next to 7 (173)
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    c) Another way to convert to binary is to simply keep dividing by two
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    <quote>
    If a number is even, dividing it by 2 will have 0 remainder.
    If it is odd, dividing it by 2 will have a 1 remainder. Dividing
    by 2 is so easy that I, at least, can just write down a column of
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    459 1
    229 1
    114 0
    57 1
    28 0
    14 0
    7 1
    3 1
    1 1
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    </quote>
    d) A final useful tip mentioned: as long as x is not zero,
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    scripting language should probably be able to handle the calculations.
    (Here's a quick sample:)
    |$ python
    |Python 2.2.2 (#1, Dec 31 2002, 12:24:34)
    |[GCC 3.2 20020927 (prerelease)] on cygwin
    |Type "help", "copyright", "credits" or "license" for more information.
    |>>> 12 & 14
    |12
    |>>> 12 | 14
    |14
    |>>> 12 ^ 14
    |2
    |>>> ~12
    |-13
    |>>> myResult = 0x5422CA66 & 458
    |>>> myResult
    |66
    |>>> hex(myResult)
    |'0x42'
    (hey, it wouldn't be a LUG without a little evangelism, would it? :)
    Much thanks to Bill, Jason, Kevin, and YATArchivist.
    Erik

  • Does SCWCD ask question about J2SE

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    Thanks

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  • Doubt about the concept of HashSet and LinkedHashSet

    I read one of the SCJP 6 exam book, when talking about Set, it gives one definition:
    When using HashSet or LinkedHashSet, the objects you add to them must override hashCode(). If they don't override hashCode(), the default Object. hashCode() method will allow multiple objects that you might consider "meaningfully equal" to be added to your "no duplicates allowed" set.
    What I am getting confused is that IF the objects we add to them must override hashCode(), we must override equals() method also! Isn't it?
    Edited by: roamer on 2009?10?23? ??10:29

    jverd wrote:
    endasil wrote:
    When using HashSet or LinkedHashSet, the objects you add to them must override hashCode(). If they don't override hashCode(), the default Object. hashCode() method will allow multiple objects that you might consider "meaningfully equal" to be added to your "no duplicates allowed" set.This really is completely wrong. Duplicates being added to your set has nothing to do with not overriding hashCode, and everything to do with not overriding equals. No, if you override equals but not hashCode, you can get dupes. That is, two items that your equals method says are equal can make it into the Set.Sorry, to clarify, what I meant by that was that to avoid duplicates of meaningfully equal objects, you must override equals, and overriding hashCode is just a consequent of overriding equals (to maintain the equal objects have equal hashcodes invariant). Overriding hashCode alone will not, and cannot, prevent meaningfully equal duplicates if you don't first define meaningfully equal by overriding equals.
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    2) If you have any other definition of meaningfully equal that you want enforced in a Set, you must override equals to correspond to that definition.
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    That's why I was saying that the statement was completely wrong. It's taking a backwards approach.
    By not overriding equals, you're saying that no two separate instances can be meaningfully equal, so the default hashCode is fine. Yeah, I kind of figured for the statement to be meaningful, it is assumed that you have already overridden equals. Otherwise what's the point of even mentioning something two objects "meaningfully equal"?All this talk of "meaningfully equal" makes me think the opposite: that they haven't overridden equals, or haven't discussed it yet. Otherwise you could just say that the objects are equal, or are equal according to equals().
    I would question the authority of the source that said that.I would quesiton the source's ability to express himself clearly. :-)Yeah, maybe that's all it is. In that case, I suggest (to the OP) reading [Effective Java, Chapter 3|http://java.sun.com/developer/Books/effectivejava/Chapter3.pdf]. Best source I've seen for beginners to make sense of all this.
    Edited by: endasil on 26-Oct-2009 1:11 PM

  • What are the advantages of SCJP??

    I know SCJP is a certificate acknoledging your Java programming level.
    But, is there any realistic advantage??
    Secondly, I see there are slightly different versions of SCJP, for eg, 1.2, or 1.4 platform. What makes them different??
    Apreciate any response

    I know SCJP is a certificate acknoledging your Java
    programming level.Kind of. It says nothing about the quality of the code you'll write.
    But, is there any realistic advantage??Depends. It seems that in some parts of the world, the recruiters value it. In other parts of the world, it's just a sheet of paper.
    Secondly, I see there are slightly different versions
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    Hm. There's a SCJP for Java 1.3, one for 1.4, one for Java 1.5... I wonder where's the difference, too.
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