Quirino Gervacio

how were questions are they hard? - if you can successfully pass K&B's 6.0 mock exams, i'm sure you'll pass the real exam.

How many drag and drop question did you get ? - i forgot. trust me it's not a big deal. although it's true that when you already answered the drag and drop questions, answering it again will erase your previous answer.

Which mock exam have you done ? - voodoo (1.x-1.4), inquisition (5.0-6.0) and K&B 6.0 mock exams and this. take as many mock exams as you can to get the feel of the exam.

Which is similar to real exam ? - it's true (what they said), the real exam is easy compared to mock exams.

3 hours is more than enough. drink a lot of water and just be

10/15/2008 09:00AM - 12:00PM



[ October 15, 2008: Message edited by: Quirino Gervacio ]

Gabriel is right, there's really no getBidValue method in the class Bridge in Self-Test Answers.

But in Sef-Test Questions, there's no getBidValue method in the code.

Compare the code of STQ and STA question #7, line 12

Hi Sanjeev,

How 'bout call it a.. hmmm

Method local anonymous class?

or

meloanclass

Have fun with Java!

Wow. Judging by the number of resources you've studied and acquired, the exam must be really hard.

In my case, I'm also working full-time (8am-5pm). I bought S&B SCJP6. If I'm not wrong, I received the book on 1st week of September. My inconsistent study hours is 7pm onwards (mostly 2 hours only) on weekdays and on weekends, it all depends if I'm in the mood. If not, I'm out with my girlfriend or just playing Warcraft on GArena.

I think what scares you is that, you had so much preparations and it doesn't make sense for you to fail or even get <90% in this exam. Stick to your promise, take one last pass, drink, get laid, have fun then schedule your exam.

If you passed then got 95%+

If not, at least you tried... then re-take
[ August 17, 2008: Message edited by: Quirino Gervacio ]

Hi Michael,

I checked the code and it compiles. The difference is that the code in the book, the class Othello is a member of GeoCache. But the code you posted, Othello is not a member of GeoCache.

I hope this helps.

Hi Bob,

I modified your code a bit
package scjpjavaranchhelp.august2008; public class BobProblem1<T>{ public static void main(String args[]){ new BobProblem1().meth(new Integer(42)); // class java.lang.Integer new BobProblem1().meth(new Short((short) 42)); // class java.lang.Short new BobProblem1().meth(new Object()); // class java.lang.Object new BobProblem1().meth(1); // class java.lang.Integer new BobProblem1().meth(false); // class java.lang.Boolean new BobProblem1<Object>().meth(new Integer(42)); // class java.lang.Integer new BobProblem1<Object>().meth(new Short((short) 42)); // class java.lang.Short new BobProblem1<Object>().meth(new Object()); // class java.lang.Object new BobProblem1<Object>().meth(1); // class java.lang.Integer new BobProblem1<Object>().meth(false); // class java.lang.Boolean new BobProblem1<Integer>().meth(new Integer(42)); // class java.lang.Integer //new BobProblem1<Integer>().meth(false); Compile error } public void meth(T type){ System.out.println(type.getClass()); } }

I think not specifying the type parameters means that the type defaults to Object.

i'm not sure here. what do you think?
[ August 17, 2008: Message edited by: Quirino Gervacio ]

Hi Gabriel,

I think it's OK since Beagle is also a Dog.

Or maybe the author is trying to tell us (in the real world) that not all Beagles are pet-able.

Hi,

public class STQ3 { public static void main(String[] args){ Short s = 2; Integer i = 1; System.out.println(s >= i); // true //Compile error: incomparable types: java.lang.Short and java.lang.Integer System.out.println(s == i); } }

Why? >= and == are both comparison operators right?

I think ARIJIT is right (if we're on the same boat)

the algorithm could be:
if there is at least ONE break in this switch (whether this break is reachable or not) then any statement after this switch is reachable else error unreachable statement end if
but for

if(true){}
else{}

may be the compiler here assumes that you know what you are doing even though us, humans can tell that the else will never get executed :-)

and for
int i;
if(false){
i=0;
}
else{}
System.out.print(i); //(1) variable i might not have been initialized
the compiler can now easily tell that no statement above that initializes i and would be very much happy to yell the error.

but if you initialize i in else (1) will be OK. maybe because the compiler
can also tell that i is now initialized. it doesn't really matter whether the statement the initialized it is reachable, as long as there is an assignment operation in i, the compiler is happy.

and of course, i could be wrong here.

i hope this helps.
[ August 11, 2008: Message edited by: Quirino Gervacio ]

Hi Ashok,

I would love to but I'm not sure if I'm allowed to send copy of the e-book version of the book.

Perhaps we could ask the moderator of this thread.

Hi to all

Here's my analysis, please correct me if I'm wrong

The default is always the last choice. I noticed that when I change iConst value to 10 and change the code to

public class AnkitProblem1{ public static void main(String[] args){ int iVal; if(args.length == 0){ iVal = 10; } else{ iVal = 10; } //(1)compiler is able to determine that iVal is set. final int iConst = 10; switch(iVal){ //System.out.println("Never Reached"); case 15: System.out.println("Never Displayed"); return; default: System.out.println("Default"); return; case iConst: System.out.println("Displayed"); return; case 30: System.out.println("Never Reached"); //(2) //return; //(3) } System.out.println("Never Reached"); //(4) } }

The output is Displayed even though default comes first before case iConst:

But in case 30:, i believe that the compiler is able to determine that it is the last case of this switch. The compiler can
tell that any code after the return of case 30: will be unreachable.

Now notice if the code is changed to

public class AnkitProblem1{ public static void main(String[] args){ int iVal; if(args.length == 0){ iVal = 10; } else{ iVal = 10; } //(1)compiler is able to determine that iVal is set. final int iConst = 10; switch(iVal){ //System.out.println("Never Reached"); case 15: System.out.println("Never Displayed"); return; case iConst: System.out.println("Displayed"); return; case 30: System.out.println("Never Reached"); //(2) return; //(3) default: System.out.println("Default"); return; } System.out.println("Never Reached"); // Compile error, unreachable statement } }

again, the compiler know that the default is the last for this switch. No case statment left.

I hope this helps.
[ August 11, 2008: Message edited by: Quirino Gervacio ]

Alas!!! i finally got it!

The answer is really one object eligible for GC and that object is a1.

I forgot that a1.b1 is misleading. It is actually Alpha.b1

Many thanks!

Object o = arr;
is valid

Notice when you perform a dot operation in arr, you'll be able to use the methods of the Object class.

Not
Object[] o = arr
Because arr is not an array of object.

Also, just like Steve posted, there is no Arrays.sort(boolean[])
The API is also your best friend. :-)

I hope this helps
[ August 10, 2008: Message edited by: Quirino Gervacio ]

From Sierra and Bates SCJP6 Study Guide
Chapter 3: Self Test, Question 11, Page 275

class Beta {} class Alpha { static Beta b1; Beta b2; } public class Tester { public static void main(String args[]){ Beta b1 = Beta(); Beta b2 = Beta(); Alpha a1 = new Alpha(); Alpha a2 = new Alpha(); a1.b1 = b1; a1.b2 = b1; a2.b2 = b2; a1 = null; b1 = null; b2 = null; //Line 16 do stuff } }
According to the book, when line 16 is reached, there is only one object eligible for garbage collection. I really don't know why. I always come up with 2 objects eligible for garbage collection which are the object refered to by b1 and a1. Please help.