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2009-04-20T19:41:00.006+05:30

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TIPS

TIPS TO CRACK SCJP

2009-02-11T00:58:00.002+05:30

1. THE BOOK:
The best reference for SCJP is Head First Java, 2nd Edition.
This books gives all the inside and out of java which is required for SCJP examination.

Herbert Schildt is a good book in general, but is not a good reference for SCJP and many of the texts in this book are not correct.
So in my opinion Head First Java, 2nd Edition is the Bible for SCJP.

2. SCJP DUMPS
SCJP Dumps are set of mock and previous year papers. Previous year questions are a must.
If one understands and masters all the questions, one can easily score in the range of 75 -100%.
These questions often get repeated and if one goes through the whole set, one is bound to get around 60-80% questions in the test from previous year papers.

3.TEST
First of all, make sure that you mark 'novice' in the survey taken by sun prior to the test(SCJP).
Don't ever mark 'intermediate' or 'expert' because the paper will become quite difficult and might be difficult to pass even.

Secondly, 'Drag and Drop Questions'- these questions have to be attempted very carefully.
Make sure you check before confirming your answer because after confirming if you try to check your answer,it gets refreshed, and your answers disappear.
So better technique is to try and check in the first attempt itself.

SCJP MOCK QUESTIONS - 2

2007-12-31T15:34:00.000+05:30

Question 2

Given:
10. public class Bar {
11.static void foo(int...x) {
12. // insert code here
13. }
14. }
Which two code fragments, inserted independently at line 12, will allow
the class to compile? (Choose two.)
A. foreach(x) System.out.println(z);
B. for(int z : x) System.out.println(z);
C. while( x.hasNext()) System.out.println( x.next());
D. for( int i=0; i< x.length; i++ ) System.out.println(x[i]);

Answer: BD

Answer:

type... x is used in function arguments where type can be any data type
for example:
int... x

which can take variable arguments
for example a function named var having declarations

var(int... x)

we can pass var(1); , var(1,1); , var(1,2,3);

so it can take 1 to any number of integers as arguments

int... x here x is just like an array and we can iterate over its elements using simple and enhanced for loop

other options are invalid

foreach is a method which is not defined
And hasnext() is used to iterate over collections(sets,lists,maps)

Hence the answer is B,D

SCJP MOCK QUESTIONS - 1

2007-12-30T10:14:00.000+05:30

QUESTION

Given:
11. public interface Status {
12. /* insert code here */ int MY_TRUE_VALUE = 10;
13. }
Which three are valid on line 12? (Choose three.)
A. final
B. static
C. native
D. public
E. private
F. abstract
G. protected

ANS - ABD

In Interfaces we can define constants for example:

interface you
{ int i=10;
}

now in interfaces every variable is implicitly public static final
Therefore, though we define int i=10 but for the compiler it is
public static final int i=10;

Now you can even define like this way :

final int i=10;
static int i=10;
public int i=10;
public static int i=10;
public final int i=10;
static final int i=10;
public static final int i=10;

All these declarations are valid but for the compiler all these declarartions lead to the same declaration - public static final int i=10;

Invalid Declarations:

native - can be only applied to methods
private - variables in interfaces are public
protected - variables in interfaces are public
abstract - As the variables are implicitly final and we know final and abstract cannot be together, therefore it is invalid.

CASTING - II

2007-12-25T10:56:00.000+05:30

Casting to (byte)

Within the range:

A byte is signed and can hold a range of numbers from -128 to 127. If the value of the widest type is within this range, conversion won't produce unexpected results.

Example:

int a = -128;

byte x = (byte)a;

float a = -128.0f;

byte x = (byte)a;

Result in both cases: -128

Outside the range but within the signed byte range:

If the value is between 128 and 255, it will be converted to binary and then to the byte decimal representation of that binary pattern. In fact, this bit-level interpretation always occurs, but you have to be conscious about it for this special case.

Example:

int a = 128;

byte x = (byte)a;

Result: -128

The bit pattern for 128 is 10000000, but 10000000 is considered to be a signed byte. Thus 10000000 is equal to -128. The next binary number, 10000001, equals to -127. If byte was unsigned, as in C/C++, the decimal value of 1000001 would be 129.

Example:

int a = 129;

byte x = (byte)a;

Result: -127

Outside the signed byte range:

If the value is greater than 255 or lower than -128, the lower byte of the value is kept and the rest is just thrown away.

Example #1:

int a = 257;

byte x = (byte)a;

Result: 1

257 = [00000000] [00000000] [00000001] [00000001]

32-bits int value

1 = [00000001]

8-bits byte value

Example #2:

int a = -135;

byte x = (byte)a;

Result: 121

-135 = [11111111] [11111111] [11111111] [01111001]

32-bits int value

121 = [01111001]

8-bits byte value

Casting to (char)

Within the range:

A char is 16-bits wide unsigned type that holds values between 0 and 65535. Conversion will perform as expected if the value is within the valid range.

Example:

int a = 65535;

char x = (char)a;

Result: 65535

Outside the range or negative:

If the value is outside the range because it is lower than 0 or greater than 65535, then the lower 2 bytes will be kept.

Example #1:

int a = 65539;

char x = (char)a;

Result: 3

65539 = [00000000] [00000001] [00000000] [00000011]

32-bits int value

3 = [00000000] [00000011]

16-bits char value

Example #2:

int a = -1;

char x = (char)a;

Result: 65535

-1 = [11111111] [11111111] [11111111] [11111111]

32-bits int value

65535 = [11111111] [11111111]

16-bits char value

Casting to (short) and other signed integer values

Values between -32768 and 32767 are converted flawlessly as they are within the valid range. If the value is lower or greater, the lower 2 bytes of the value will be kept to conform a short value. The behaviour is the same as byte casting.

Other integer values will also behave as expected according to what we've seen at the byte examples.

Casting floats or doubles to narrower types

On some programming languages, conversion from floating-point numbers to decimals "round" the value. This is not the Java case, the integer part is kept and the rest is thrown away.

Example:

double a = 1.9999999;

int x = (int)a;

Result: 1

Object Reference Conversion

Object reference conversion takes place in:

* Assignment
* Method call
* Casting

(There is no arithmetic promotion)

Assignment

Object reference assignment conversion happens when you assign an object reference value to a variable of a different type.

There are 3 general kinds of object reference type:

* A class type, such as Button or TextField
* An interface type, such as Clonable or LayoutManager
* An array type, such as int[][] or TextArea[]
*

Conversion rules for implicit casting on this context:

OLD_Type a = new OLD_Type;

NEW_Type b = a;

CASTING

2007-12-25T10:55:00.000+05:30

Primitives

Conversion of primitive types may occur in these contexts:

* Assignment
* Method call
* Arithmetic promotion
* Explicit casting

Widening conversions: -

Your browser may not support display of this image.

Assignment

General rules for primitive assignment conversions:

* A boolean may not be converted to any other type.
* A non-boolean may be converted to another non-boolean type, provided the conversion is a widening conversion.
* A non-boolean may not be converted to another non-boolean type, if the conversion would be a narrowing conversion.

Example #1:

int i = 5;

float j = i;

Method call

Widening conversion takes place on method calls as on assignments. You can pass to a method any primitive narrower than the expected one. Implicit casting will naturally occur.

For instance:

public static void main(String args[]) {

byte x = 126;

System.out.println( DoIt(x) );

}

static String DoIt(int a) {

return "I've received an int of value "+a;

}

Result: I've received an int value of 126

The method DoIt(int a) expects an int, but you can throw a char, byte or short there, the value will be promoted to int and the method will IN FACT receive an int.

This special behavior occurs if a method to handle a narrower type hasn't been declared. If you declare a method to handle bytes, then that method will "catch" the call. This is an OOP feature called overloading.

Example:

public static void main(String args[]) {

byte x = 126;

System.out.println( DoIt(x) );

}

static String DoIt(int a) {

return "I've received an int of value "+a;

}

static String DoIt(byte a) {

return "I've received a byte of value "+a;

}

Result: I've received a byte value of 126

If the argument type if wider than expected, no implicit casting will occur and you will need to perform an explicit cast:

public static void main(String args[]) {

float x = 1.26f;

System.out.println( DoIt( (int)x ) );

}

static String DoIt(int a) {

return "I've received an int of value "+a;

}

Result: I've received an int value of 1

Last example:

public static void main(String args[]) {

char x = 'A';

System.out.println( DoIt(x) );

}

static String DoIt(int a) {

return "I've received an int of value "+a;

}

static String DoIt(byte a) {

return "I've received a byte of value "+a;

}

Result: I've received an int value of 65

As you can see, there's no method to catch char types so the value is promoted to int and caught by DoIt(int a).

Arithmetic Promotion

Arithmetic promotion happens when narrower types need to be promoted to wider types in order to make sense in an operation among other wider types.

Basic rules for binary operators and most of the unary operators:

* All arithmetic expressions are promoted to the wider of the operands.
* The promotion is at least to int, even if no int operand appears.

These rules don't apply to the unary operators: ++ and -- and assignment operators.

Example #1:

byte x = 1;

x++; // Ok. x is now equal to 2.

x = x + 1; // Error. Expression x + 1 is promoted to int.

Example #2:

byte a = 1;

byte x = 23;

x <<= a; // Ok. x is now equal to 46.

x = x << a; // Error. Expression x << a is promoted to int.

Example #3:

char a = 5;

short x = 3;

x *= a; // Ok. x is now equal to 15.

x = x * a; // Error. Expression x = x * a is promoted to int.

Example #4:

byte a = 15;

a = -a; // Error. -a is promoted to int.

a = ~a; // Error. ~a is promoted to int.

Example #5:

float a = 1.0f;

int b = 15;

int x = a * b; // Error. Expression is promoted to float.

int x = (int)(a*b); // Ok. We cast the float result back to int.

Primitives and Casting

Casting means explicitly telling Java to make a conversion. A casting operation may widen or narrow its argument. To cast a value, you need to precede it with the name of the desired type enclosed within parentheses:

byte x = 15;

int r = (int)(x * 3);

Booleans cannot be casted. Don't bother with them, stuff like this doesn't work:

byte a = 1;

boolean status = a;

Narrowing runs the risk of loosing information, in fact, many times you know that you are going to loose information and it is important to know which part information is going to be loosen and naturally, what information will be kept.

Java Video Tutorials

2007-12-23T14:27:00.000+05:30

Are you new to java? Then these video tutorials will help you tke a step forward towards learning java programming.

This video gives an example of a very simple program to make a new java file, compile and run and print a cutom text on the screen.

The second video is a tutorial to start using java in Eclipse environment.

ARRAYS - II

2007-12-15T11:38:00.001+05:30

Intermediate Arrays

Anonymous arrays

Java 2 added anonymous arrays, which allow you to create a new array of values anywhere in the program, not just in an initialization in a declaration.

// This anonymous array style can also be used in other statements.

String[] days = new String[] {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

You can also use anonymous array syntax in other parts of the program. For example,

// Outside a declaration you can make this assignment.

x = new String[] {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

You must be careful not to create these anonymous arrays in a loop or as local variables because each use of new will create another array.

Dynamic allocation

Because arrays are allocated dynamically, the initialization values may arbitrary expresssions. For example, this call creates two new arrays to pass as parameters to drawPolygon.

g.drawPolygon(new int[] {n, n+45, 188}, new int[] {y/2, y*2, y}, 3);

C-style array declarations

Java also allows you to write the square brackets after the variable name, instead of after the type. This is how you write array declarations in C, but is not a good style for Java. C declaration syntax can get very ugly with part of the declaration before the variable, and part after. Java has a much nicer style where all type information can be written together without the use of a variable, and there are times when only the Java notation is possible to use.

int[] a; // Java style -- good

int a[]; // C style -- legal, but not Java style

Converting between arrays and Collections data structures

[TO DO]

Common array problems

Some common array programming mistakes are:

Forgetting that array subscripts start with zero.
Writing a.length() instead of a.length. The length() method is used with Strings, not arrays.
Declaring an array with a size. Eg, int[100] a; instead of int[] a = new int[100];.

Library methods for arrays

Static methods for manipulating arrays are available in the java.util.Arrays class.

Arrays.asList()	Returns a List based on the array.
Arrays.toString()	Returns a readable form of the array.
Arrays.binarySearch()	Performs binary serach of a sorted array.
Arrays.equals	Compares two arrays for equality.
Arrays.fill()	Fills entire array or subrange with a value.
Arrays.sort()	Sorts an array.

In addition there is System.arrayCopy() method.

Anonymous arrays have to be assigned to the variables at the time of declaration of varaible itself. Assigning to the variable after variable declaration is not possible
ie.,
int[] i = {1,2,3} is fine.
But,
int[] i;
i = {1,2,3} is compiler error.
This restriction at the time of method invocation is valid, since we don't know the type of array(overloading issue). But for assignment the left side variable type is known. So, why should it be not possible to cast the anonymous array {1,2,3} to int[] and assign to variable.

i=i++ produces the output "0″ instead of "1″.

The code
int i = 0;
i = i++;
System.out.println(i);
produces the output "0″ instead of "1″.

"i = i++" roughly translates to

int oldValue = i;
i = i + 1;
i = oldValue;

Array and ArrayList

In a arrayList of object, any types of object(mixed) can be put. However in an array of objects, once initialized to a particular object type, other type of objects cannot be put.

Thus,

ArrayList a = new ArrayList();
a.add("krishna");
a.add(5);

is allowed and will run fine.

However,

Object[] obj = new String[5];
obj[0] = new Integer(1);

will throw ArrayStoreException(is a RunTimeException).

With generics, arraylist is also not allowed to store Integer if it is declared as ArrayList of String. Thus, with generics,
ArrayList<String> a = new ArrayList<String>();
a.add("krishna");
a.add(5); — will be compiler error(unlike array which is at rutime).

But with generics also, one more difference is array allows polymorphic assignment., i.e.,

Number[] n = new Integer[5];

However, in case of generics, polymorphic assignment holds good only for the base

type and not for the generic type. i.e.,

List<Integer> l = new ArrayList<Integer>; — for base type hold good

ArrayList<Number> n = new ArrayList<Integer>; — not allowed for generic type

ARRAYS - I

2007-12-15T11:36:00.001+05:30

Arrays

Java arrays are similar to ideas in mathematics

An array can store many similar values in memory. Each value can be accessed by specifying a subscript or index. "Array" in Java means approximately the same thing as array, matrix, or vector does in math.

Unlike math, you must declare the array and allocate a fixed amount of memory for it.

Declaring an array

An array variable is like other variables -- you must declare it, which means you must declare the type of elements that are in an array. All elements must be the same type. Write the element type name, then "[]", then the name of the array variable. The declaration allocates only enough space for a reference to an array (typically 4 bytes), but doesn't create the actual array object.

String[] args; // args is an array of Strings

int[] scores; // scores is an array of ints

JButton[] bs; // bs is an array of JButtons

No size in declaration. Unlike some languages, never put the size of the array in the declaration because an array declaration specifies only the element type and the variable name.

Allocate an array object with new

Create an array using new. This example creates an array of 100 int elements, from a[0] to a[99].

int[] a; // Declare a to be an array of ints

a = new int[100]; // Allocate an array of 100 ints

These are often combined in one line.

int[] a = new int[100]; // Declare and allocate.

Subscripts

Subscripts are enclosed in square brackets []. x_i in mathematics is x[i] in Java, and is pronounced "x-sub-i".

Subscript ranges always start at zero because Java came largely from C++, which had a good reason for using zero (pointer arithmetic on arrays). It isn't the way that humans normally count; you'll just have to live with it.

Java always checks subscript legality to be sure the subscript is >= 0, and less than the number of elements in the array. If the subscript is outside this range, Java throws ArrayIndexOutOfBoundsException. This is far superior to the behavior of C and C++, which allow out of range references. Consequently, Java programs are far less susceptible to bugs and security flaws than C/C++ programs.

Length of an array

Each array has a constant (final) instance variable that has its length. You can find out how many elements an array can hold by writing the array name followed by .length. In the previous example, a.length would be 100. Remember that this is the number of elements in the array, one more than the maximum subscript.

Java idiom for looping over an array

The most common use of .length is in a for loop test condition. For example, the variable i will go over the entire range of subscripts of the array a.

for (int i=0; i < a.length; i++) {

. . .

}

If you only need to reference the value of each of the elements, you can use the somewhat simpler Java 5 for loop, which keeps track of the index and assigns successive values to a variable (v in this example).

for (int v : a) {

. . .

}

Example Version 1 -- Adding all elements of an array

These statements create an array and put 1000 random values in it. The second loop adds all 1000 elements. It would have been better to add them in the first loop, but writing it this way allows two examples of loops.

int[] a; // Declare an array of ints

a = new int[1000]; // Create an array of 1000 ints, a[0]...a[999]

//... Assign random values to each element.

for (int i=0; i<a.length(); i++) {

a[i] = (int)(Math.random() * 100000); // Random number 0-99999

}

//... Add all values in the array.

int sum = 0; // Start the total sum at 0.

for (int i=0; i<a.length; i++) {

sum = sum + a[i]; // Add the next element to the total

}

Example Version 2 -- Adding all elements of an array in Java 5

This is the the same as above, but uses the Java 5 "for each" loop to do the sum, which frees you from using an index if you simply need to get all successive values. This kind of loop only gets the values, so it couldn't have been used to set the values in the first loop above.

. . .

int sum = 0; // Start the total sum at 0.

for (int v : a) {

sum = sum + v; // Add the next element to the total

}

Initial array element values -- zero/null/false

When an array is allocated (with new), all elements are set to an initial value. The initial value is 0 if the element type is numeric (int, float, ...), false for boolean, and null for all object types.

Array Initialization

When you declare an array, you can can also allocate a preinitialized array object in the same statement. In this case, do not give the array size because Java counts the number of values to determine the size. For example,

// Java 1.0 style -- shorter, but can be used ONLY IN DECLARATIONS

String[] days = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

Array variables are references to arrays

When you declare an array variable, Java reserves only enough memory for a reference (Java's name for an address or pointer) to an array object. References typically require only 4 bytes. When an array object is created with new, a reference is returned, and that reference can then be assigned to a variable. When you assign one array variable to another, only the reference is copied. For example,

int[] a = new int[] {100, 99, 98};

int[] b;

// "a" points to an array, and "b" doesn't point to anything

b = a; // Now b referes to the SAME array as "a"

b[1] = 0; // Also changes a[1] because a and b refer to the same array.

// Both a and b refer to same array with value {100, 0, 98}

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2007-12-14T13:56:00.000+05:30

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JAVA DUMPS 19 (SCJP)

2007-12-14T10:38:00.001+05:30

QUESTION NO: 6

Which statement is true?

A. Memory is reclaimed by calling Runtime.gc().

B. Objects are not collected if they are accessible from live threads.

C. Objects that have finalize() methods are never garbage collected.

D. Objects that have finalize() methods always have their finalize() methods called before

the program ends.

E. An OutOfMemory error is only thrown if a single block of memory cannot be found

that is large enough for a particular requirement.

Answer: B

QUESTION NO: 7

Given:

1. class A {

2. A() { }

3. }

5. class B extends A {

6. }

Which two statements are true? (Choose two)

A. Class B's constructor is public.

B. Class B's constructor has no arguments.

C. Class B's constructor includes a call to this().

D. Class B's constructor includes a call to super().

Answer: B, D

QUESTION NO: 8

Given:

11. int i = 1,j = 10;

12. do {

13. if(i>j) {

14. break;

15. }

16. j--;

17. } while (++i <5);

18. System.out.println("i =" +i+" and j = "+j);

What is the result?

A. i = 6 and j = 5

B. i = 5 and j = 5

C. i = 6 and j = 4

D. i = 5 and j = 6

E. i = 6 and j = 6

Answer: D

QUESTION NO: 9

Which statement is true?

A. Assertions can be enabled or disabled on a class-by-class basis.

B. Conditional compilation is used to allow tested classes to run at full speed.

C. Assertions are appropriate for checking the validity of arguments in a method.

D. The programmer can choose to execute a return statement or to throw an exception if

an assertion fails.

Answer: A

QUESTION NO: 10

You want a class to have access to members of another class in the same package. Which

is the most restrictive access that accomplishes this objective?

A. public

B. private

C. protected

D. transient

E. default access

Answer: E

JAVA DUMPS 18 (SCJP)

2007-12-14T10:36:00.001+05:30

QUESTION NO: 1

Given:

1. public class Test {

2. public static void main(String args[]) {

3. class Foo {

4. public int i = 3;

5. }

6. Object o = (Object)new Foo();

7. Foo foo = (Foo)o;

8. System.out.println("i = " + foo.i);

9. }

10. }

What is the result?

A. i = 3

B. Compilation fails.

C. A ClassCastException is thrown at line 6.

D. A ClassCastException is thrown at line 7.

Answer: A

QUESTION NO: 2

Which two cause a compiler error? (Choose two)

A. float[] = new float(3);

B. float f2[] = new float[];

C. float[] f1 = new float[3];

D. float f3[] = new float[3];

E. float f5[] = { 1.0f, 2.0f, 2.0f };

F. float f4[] = new float[] { 1.0f. 2.0f. 3.0f};

Answer: A, B

QUESTION NO: 3

Given:

11. int i =1,j =10;

12. do {

13. if(i++> --j) {

14. continue;

15. }

16. } while (i <5);

17. System.out.println("i = " +i+ "and j = "+j);

What is the result?

A. i = 6 and j = 5

B. i = 5 and j = 5

C. i = 6 and j = 5

D. i = 5 and j = 6

E. i = 6 and j = 6

Answer: D

QUESTION NO: 4

Given:

1. class Test {

2. private Demo d;

3. void start() {

4. d = new Demo();

5. this.takeDemo(d);

6. }

8. void takeDemo(Demo demo) {

9. demo = null;

10. demo = new Demo();

11. }

12. }

When is the Demo object, created on line 3, eligible for garbage collection?

A. After line 5.

B. After line 9.

C. After the start() method completes.

D. When the takeDemo() method completes.

E. When the instance running this code is made eligible for garbage collection.

Answer: E

QUESTION NO: 5

Given:

1. interface Animal {

2. void soundOff();

3. }

5. class Elephant implements Animal {

6. public void soundOff() {

7. System.out.println("Trumpet");

8. }

9. }

10.

11. class Lion implements Animal {

12. public void soundOff() {

13. System.out.println("Roar");

14. }

15. }

16.

17. class Alpha1 {

18. static Animal get( String choice ) {

19. if ( choice.equalsIgnoreCase( "meat eater" )) {

20. return new Lion();

21. } else {

22. return new Elephant();

23. }

24. }

25. }

Which compiles?

A. new Animal().soundOff();

B. Elephant e = new Alpha1();

C. Lion 1 = Alpha.get("meat eater");

D. new Alpha1().get("veggie").soundOff();

Answer: D

JAVA DUMPS 17 (SCJP)

2007-12-11T16:09:00.001+05:30

QUESTION NO: 92

Given:

1. abstract class AbstractIt {

2. abstract float getFloat();

3. }

4. public class AbstractTest extends AbstractIt {

5. private float f1 = 1.0f;

6. private float getFloat() { return f1; }

7. }

What is the result?

A. Compilation succeeds.

B. An exception is thrown.

C. Compilation fails because of an error at line 2.

D. Compilation fails because of an error at line 6.

Answer: D

QUESTION NO: 93

Which four can be thrown using the throw statement? (Choose four)

A. Error

B. Event

C. Object

D. Throwable

E. Exception

F. RuntimeException

Answer: A, D, E, F

QUESTION NO: 94

What produces a compiler error?

A. class A {

public A(int x) {}

}

B. class A {

}

class B extends A {

B() {}

}

C. class A {

A() {}

}

class B {

public B() {}

}

D. class Z {

public Z(int) {}

}

class A extends Z {

}

Answer: D

QUESTION NO: 95

Given:

11. for( int i = min; i <max; i++) {

12. System.out.println(i);

13. }

If min and max are arbitrary integers, what gives the same result?

A. init i = min;

while( i < max ) {

}

B. int i = min;

System.out.println(i++);

} while( i< max );

C. for (int i=min; i<max; System.out.println(++I));

D. for (int i=; i++<max; System.out.println(i));

Answer: B

JAVA DUMPS 16 (SCJP)

2007-12-09T10:02:00.001+05:30

QUESTION NO: 85

Given:

12. float f[][][] = new float[3][][];

13. float f0 = 1.0f;

14. float[][] farray = new float[1][1];

What is valid?

A. f[0] = f0;

B. f[0] = farray;

C. f[0] = farray[0];

D. f[0] = farray[0][0];

Answer: B

QUESTION NO: 86

Given:

11. for (int i =0; i < 4; i +=2) {

12. System.out.print(i + "");

13. }

14. System.out.println(i);

What is the result?

A. 0 2 4

B. 0 2 4 5

C. 0 1 2 3 4

D. Compilation fails.

E. An exception is thrown at runtime.

Answer: D

QUESTION NO: 87

Given:

12. void start() {

13. A a = new A();

14. B b = new B();

15. a.s(b);

16. b = null;

17. a = null;

18. System.out.printIn("start completed");

19. }

When is the B object, created in line 14, eligible for garbage collection?

A. After line 16.

B. After line 17.

C. After line 18 (when the methods ends).

D. There is no way to be absolutely certain.

E. The object is NOT eligible for garbage collection.

Answer: C

QUESTION NO: 88

Given:

1. public class Exception Test {

2. class TestException extends Exception {}

3. public void runTest() throws TestException {}

4. public void test() /* Point X */ {

5. runTest();

6. }

7. }

At Point X on line 4, which code is necessary to make the code compile?

A. No code is necessary.

B. throws Exception

C. catch ( Exception e )

D. throws RuntimeException

E. catch ( TestException e)

Answer: B

QUESTION NO: 89

Given:

11. int i = 0;

12. while (true) {

13. if(i==4) {

14. break;

15. }

16. ++i;

17. }

18. System.out.println("i="+i);

What is the result?

A. i = 0

B. i = 3

C. i = 4

D. i = 5

E. Compilation fails.

Answer: C

QUESTION NO: 90

Given:

11. try {

12. int x = 0;

13. int y = 5 / x;

14. } catch (Exception e) {

15. System.out.println("Exception");

16. } catch (ArithmeticException ae) {

17. System.out.println("Arithmetic Exception");

18. }

19. System.out.println("finished");

What is the result?

A. finished

B. Exception

C. Compilation fails.

D. Arithmetic Exception

Answer: C

QUESTION NO: 91

Given:

1. public class Test { }

What is the prototype of the default constructor?

A. Test()

B. Test(void)

C. public Test()

D. public Test(void)

E. public void Test()

Answer: A

JAVA DUMPS 15 (SCJP)

2007-12-09T09:59:00.001+05:30

QUESTION NO: 78

Given:

11. public class Test {

12. public void foo() {

13. assert false;

14. assert false;

15. }

16. public void bar(){

17. while(true){

18. assert false;

19. }

20. assert false;

21. }

22. }

What causes compilation to fail?

A. Line 13

B. Line 14

C. Line 18

D. Line 20

Answer: D

QUESTION NO: 79

Which statement is true?

A. Programs will not run out of memory.

B. Objects that will never again be used are eligible for garbage collection.

C. Objects that are referred to by other objects will never be garbage collected.

D. Objects that can be reached from a live thread will never be garbage collected.

E. Objects are garbage collected immediately after the system recognizes they are

eligible.

Answer: D

QUESTION NO: 80I

n which two cases does the compiler supply a default constructor for class A? (Choose

two)

A. class A {

}

B. class A {

public A() {}

}

C. class A {

public A(int x) {}

}

D. class Z {}

class A extends Z {

void A() {}

}

Answer: A, D

QUESTION NO: 81

Given:

1. public class ReturnIt {

2. return Type methodA(byte x, double y) {

3. return (short)x / y * 2;

4. }

5. }

What is the narrowest valid returnType for methodA in line2?

A. int

B. byte

C. long

D. short

E. float

F. double

Answer: F

QUESTION NO: 82

Given:

1. public class Outer{

2. public void someOuterMethod() {

3. // Line 3

4. }

5. public class Inner{}

6. public static void main( String[]argv ) {

7. Outer o = new Outer();

8. // Line 8

9. }

10. }

Which instantiates an instance of Inner?

A. new Inner(); // At line 3

B. new Inner(); // At line 8

C. new o.Inner(); // At line 8

D. new Outer.Inner(); // At line 8

Answer: A

QUESTION NO: 83

What allows the programmer to destroy an object x?

A. x.delete()

B. x.finalize()

C. Runtime.getRuntime().gc()

D. Explicitly setting the object's reference to null.

E. Ensuring there are no references to the object.

F. Only the garbage collection system can destroy an object.

Answer: F

QUESTION NO: 84

Given:

11. int x = 1, y =6;

12. while (y--) {

13. x++;

14. }

15. System.out.println("x =" + x + "y =" +y);

What is the result?

A. x = 6 y = 0

B. x = 7 y = 0

C. x = 6 y = -1

D. x = 7 y = -1

E. Compilation fails.

Answer: D

JAVA DUMPS 14 (SCJP)

2007-12-09T09:56:00.001+05:30

QUESTION NO: 72

You want subclasses in any package to have access to members of a superclass. Which is

the most restrictive access that accomplishes this objective?

A. public

B. private

C. protected

D. transient

E. default access

Answer: C

QUESTION NO: 73

Given:

1. class Exc0 extends Exception { }

2. class Exc1 extends Exc0 { }

3. public class Test {

4. public static void main(String args[]) {

5. try {

6. throw new Exc1();

7. } catch (Exc0 e0) {

8. System.out.println("Ex0 caught");

9. } catch (Exception e) {

10. System.out.println("exception caught");

11. }

12. }

13. }

What is the result?

A. Ex0 caught

B. exception caught

C. Compilation fails because of an error at line 2.

D. Compilation fails because of an error at line 6.

Answer: A

QUESTION NO: 74

Given:

20. public float getSalary(Employee e) {

21. assert validEmployee(e);

22. float sal = lookupSalary(e);

23. assert (sal>0);

24. return sal;

25. }

26. private int getAge(Employee e) {

27. assert validEmployee(e);

28. int age = lookupAge(e);

29. assert (age>0);

30. return age;

31. }

Which line is a violation of appropriate use of the assertion mechanism?

A. line 21

B. line 23

C. line 27

D. line 29

Answer: A

QUESTION NO: 75

Given:

1. public class A {

2. void A() {

3. System.out.println("Class A");

4. }

5. public static void main(String[] args) {

6. new A();

7. }

8. }

What is the result?

A. Class A

B. Compilation fails.

C. An exception is thrown at line 2.

D. An exception is thrown at line 6.

E. The code executes with no output.

Answer: E

QUESTION NO: 76

Given:

1. class Bar { }

1. class Test {

2. Bar doBar() {

3. Bar b = new Bar();

4. return b;

5. }

6. public static void main (String args[]) {

7. Test t = new Test();

8. Bar newBar = t.doBar();

9. System.out.println("newBar");

10. newBar = new Bar();

11. System.out.println("finishing");

12. }

13. }

At what point is the Bar object, created on line 3, eligible for garbage collection?

A. After line 8.

B. After line 10.

C. After line 4, when doBar() completes.

D. After line 11, when main() completes.

Answer: C

QUESTION NO: 77

Given:

1. interface Beta {}

3. class Alpha implements Beta {

4. String testIt() {

5. return "Tested";

6. }

7. }

9. public class Main1 {

10. static Beta getIt() {

11. return new Alpha();

12. }

13. public static void main( String[] args ) {

14. Beta b = getIt();

15. System.out.println( b.testIt() );

16. }

17. }

What is the result?

A. Tested

B. Compilation fails.

C. The code runs with no output.

Answer: B

JAVA DUMPS 13(SCJP)

2007-12-06T17:33:00.001+05:30

Given:

1. class Super {

2. public int i = 0;

4. public Super(String text) {

5. i = 1;

6. }

7. }

9. public class Sub extends Super {

10. public Sub(String text) {

11. i = 2;

12. }

13.

14. public static void main(String args[]) {

15. Sub sub = new Sub("Hello");

16. System.out.println(sub.i);

17. }

18. }

What is the result?

A. 0

B. 1

C. 2

D. Compilation fails.

Answer: D

QUESTION NO: 68

Given:

11. int i = 1,j = 10;

12. do{

13. if (i>j) {

14. continue;

15. }

16. j--;

17. } while (++i <6);

18. System.out.println("i = " +i+" and j = "+j);

What is the result?

A. i = 6 and j = 5

B. i = 5 and j = 5

C. i = 6 and j = 4

D. i = 5 and j = 6

E. i = 6 and j = 6

Answer: D

QUESTION NO: 69

Which fragment is an example of inappropriate use of assertions?

A. assert (!(map.contains(x)));

map.add(x);

B. if (x > 0) {

} else {

assert (x==0);

}

C. public void aMethod(int x) {

assert (x > 0);

}

D. assert (invariantCondition());

return retval;

E. switch (x) {

case 1: break;

case 2: creak;

default: assert (x == 0);

Answer: C

QUESTION NO: 70

Given:

1. public class X {

2. public X aMethod() { return this;}

3. }

1. public class Y extends X {

3. }

Which two methods can be added to the definition of class Y? (Choose two)

A. public void aMethod() {}

B. private void aMethod() {}

C. public void aMethod(String s) {}

D. private Y aMethod() { return null; }

E. public X aMethod() { return new Y(); }

Answer: C, E

QUESTION NO: 71

Given:

1. public class X {

2. public static void main(String [] args) {

3. try {

4. badMethod();

5. System.out.print("A");

6. }

7. catch (Exception ex) {

8. System.out.print("B");

9. }

10. finally {

11. System.out.print("B");

12. }

13. System.out.print("D");

14. }

15. public static void badMethod() {

16. throw new Error();

17. }

18. }

What is the result?

A. ABCD

B. Compilation fails.

C. C is printed before exiting with an error message.

D. BC is printed before exiting with an error message.

E. BCD is printed before exiting with an error message.

Answer: C

ANONYMOUS CLASSES

2007-12-01T12:31:00.000+05:30

Anonymous classes can be declared to extend another class or to implement a single interface. If you declare a class that implements a single explicit interface, then it is a direct subclass of java.lang.Object.

Anonymous classes give you a convenient way to avoid having to think up trivial names for classes. They should be small and easy to understand as they do not contain descriptive names.

You cannot define any specific constructor for an anonymous inner class. This is a direct consequence of the fact that you do not specify a name for the class, and therefore you cannot use that name to specify a constructor. Anonymous Class Declarations

new Identifier() { /* class body */ }

Identifier is a class or interface name. The expression by itself isn't of much use, it returns a reference to an object that you usually assign to a reference variable of the same type of the identifier:

Identifier = new Identifier() { /* class body */ };

Of course you can also pass the resulting reference to a method:

Method(new Identifier() { /* class body *} );

Example:

class Test {

public static void main(String args[]) {

Base o = new Base() {

public void Hi() {

System.out.println("Hi!");

}

public void Bye() {

System.out.println("Bye");

}

};

o.Hi();

}

interface Base {

public void Hi();

}

Result:

Hi!

Anonymous classes should provide methods defined by the base class or the interface but not brand new methods although they can be declared without compilation errors.

Passing arguments

You can define a constructor inside a inner class, but if you provide arguments, the matching constructor of base class will be invoked:

class Test {

public static void main(String args[]) {

Base o = new Base(15) {

public void Hi() {

System.out.println("Hi!");

}

public void Bye() {

System.out.println("Bye");

}

};

o.Hi();

// o.Bye(); // ERROR.

}

class Base {

Base (int i) {

System.out.println("I've got "+i);

}

public void Hi() { }

}

Result:

I've got 15

Hi!

Initialising Anonymous classes

You can't define constructors for an anonymous class, but you can declare an initialisation block:

class Test {

public static void main(String args[]) {

Object o = new Object() {

{

System.out.println("Init");

}

};

}

Result:

Init

This feature is available to all classes, not only anonymous ones.

INNER CLASSES

2007-12-01T12:27:00.000+05:30

An inner or nested class is the same as any other class but is declared inside some other class or method. When an instance of an inner class is created, there must normally be a pre-existing instance of the outer class acting as context. An inner class and an outer class belong together; the inner class is not just another member of the outer instance.

Example #1:

class Test {

class Inner {

Inner () {

System.out.println("Hello World");

}

public static void main(String args[]) {

Test.Inner i = new Test().new Inner();

}

Result:

Hello World

Please note the special syntax used to reference the Inner class from a static context. This is just a shorter approach for this:

Test t = new Test();

Inner i = t.new Inner();

Example #2:

class Test {

String h = "Hello World";

class Inner {

Inner () {

System.out.println(h);

Bye();

}

public static void main(String args[]) {

Test.Inner i = new Test().new Inner();

}

void Bye() {

System.out.println("Good bye!");

}

Result:

Hello World

Bye

Inner classes have access to all the features of the outer class including also methods.

Access modifiers and Static Inner Classes

Inner classes may be marked with standard access modifiers (private, public, protected) (or default if no modifier is specified). Static inner classes do not have any reference to the enclosing instance. Static methods of inner classes may not access non-static features of the outer class.

Example:

class Test {

private static String h = "Hello World";

static class Inner {

static void MyMethod () {

System.out.println(h);

}

public static void main(String args[]) {

Test.Inner.MyMethod();

}

Result:

Hello World

The Inner class is an extension of the outer class, so we have access to private members.

Classes defined inside Methods

Anything declared inside a method is not a member of the class but is local to the method. Therefore, classes declared in methods are private to the method and cannot be marked with any access modifier; neither can they be marked as static. However, an object created from an inner class within a method can have some access to the variables of the enclosing method if they declare the final modifier.

Since local variables and method arguments are conventionally destroyed when their method exits, these variables would be invalid for access by inner class methods after the enclosing method exists. By allowing access only to final variables, it becomes possible to copy the values of those variables into the object itself.

Example:

class Test {

public static void main(String args[]) {

Hi("Ernest");

}

static void Hi(String name) {

final String h = "Hello World " + name;

int j = 5;

class Inner {

void MyMethod () {

System.out.println(h);

// j++; // ERROR !!!

}

Inner i = new Inner();

i.MyMethod();

}

Although the variable name entered the method as normal variable, its contents were added to a final variable.

OVERLOADING CONSTRUCTORS 2

2007-12-01T12:26:00.001+05:30

Example #5:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal (int l) {

legs = l;

System.out.println("Animal constructor - leg argument");

}

class Reptile extends Animal {

Reptile (int l) {

super(l);

System.out.println("Reptile constructor - leg argument");

}

Result:

Animal constructor - leg argument

Reptile constructor - leg argument

If the Reptile class hasn't the Reptile (int l) constructor it wouldn't be possible to initialise it with that argument as constructors are not inherited as methods. Something like this would also be invalid:

Reptile r = new Reptile();

Animal a = new Animal();

Animal() and Reptile() aren't defined. Java doesn't supply default constructors because one is already defined.

Because of this special behavior you must take special care with this() because a super() call is produced automatically:

Example #6:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal (int l) {

legs = l;

System.out.println("Animal constructor - leg argument");

}

Animal (boolean b) {

cold_blood = b;

System.out.println("Animal constructor - blood argument");

}

class Reptile extends Animal {

Reptile (int l) {

this(true); // WATCH OUT!!!

System.out.println("Reptile constructor - leg argument");

}

Reptile (boolean b) {

cold_blood = b;

System.out.println("Reptile constructor - blood argument");

}

Result:

ERROR. constructor Animal () not found

There are two ways to solve this problem, we can add a Animal() constructor, or we can make sure that the proper Animal constructor gets called, like this:

Example #7:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal (int l) {

legs = l;

System.out.println("Animal constructor - leg argument");

}

Animal (boolean b) {

cold_blood = b;

System.out.println("Animal constructor - blood argument");

}

class Reptile extends Animal {

Reptile (int l) {

this(true);

System.out.println("Reptile constructor - leg argument");

}

Reptile (boolean b) {

super(b);

System.out.println("Reptile constructor - blood argument");

}

You have to understand that each parent class will be initialised somehow and you must tell Java which constructors should be used if you don't want Java to insist on calling the default constructor all the time.

So, do not forget:

In a hierarchy of classes, Java calls the default constructor of each parent class before the body of the current constructor is executed. You cannot prevent Java from calling a constructor for each class but you can tell Java which constructor should be chosen by the parent class using the super() keyword.
Java provides a default constructor for each class if NO CONSTRUCTOR is defined by the programmer. If just one constructor is defined, Java doesn't put up a default constructor anymore.
Constructors are not inherited, you must define all the constructors again or reuse the functionality of a parent constructor using super().

OVERLOADING CONSTRUCTORS 1

2007-12-01T12:25:00.000+05:30

Overloading constructors behave just like overloading methods. When you have different constructors and want to call one version from another, you use the keyword this.

Example:

class Test {

public static void main(String args[]) {

Animal d = new Animal(true);

System.out.println("Legs: "+ d.legs);

System.out.println("Mammal: "+ d.mammal);

System.out.println("Vertebrate: "+ d.vertebrate);

}

class Animal {

int legs = 0;

boolean mammal = false;

boolean vertebrate = false;

Animal (int l, boolean m, boolean v) {

legs = l;

mammal = m;

vertebrate = v;

System.out.println("Animal constructor - all arguments");

}

Animal (boolean m) {

this(4,m,true);

System.out.println("Animal constructor - mammal argument");

}

Result:

Animal constructor - all arguments

Animal constructor - mammal argument

Legs: 4

Mammal: true

Vertebrate: true

The keyword this must also be the first sentence of the constructor. Thus you can't use super() and this() at the same time in the same constructor. However, super() is implicitly called even if you don't use it because the parents of the class are always initialised first and before the body of the constructor.

Example #1:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

}

class Reptile extends Animal {

Reptile (int l) {

legs = l;

System.out.println("Reptile constructor - leg argument");

}

Result:

Reptile constructor - leg argument

The compiler calls the default constructor for Animal and then the argument version of Reptile. Let's define our own default constructor just to be sure.

Example #2:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal() {

System.out.println("Animal constructor - default");

}

class Reptile extends Animal {

Reptile (int l) {

legs = l;

System.out.println("Reptile constructor - leg argument");

}

Result:

Animal constructor - default

Reptile constructor - leg argument

Oops! We should also be able to define the number of legs for all animals, not only reptiles.

Example #3:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal() {

System.out.println("Animal constructor - default");

}

Animal (int l) {

legs = l;

System.out.println("Animal constructor - leg argument");

}

class Reptile extends Animal {

Reptile (int l) {

legs = l;

System.out.println("Reptile constructor - leg argument");

}

Results:

Animal constructor - default

Reptile constructor - leg argument

Java keeps calling the default constructor. If we take it out maybe Java will try to use the remaining method Animal (int l):

Example #4:

class Test {

public static void main(String args[]) {

Reptile r = new Reptile(4);

}

class Animal {

int legs = 0;

boolean cold_blood = false;

Animal (int l) {

legs = l;

System.out.println("Animal constructor - leg argument");

}

class Reptile extends Animal {

Reptile (int l) {

legs = l;

System.out.println("Reptile constructor - leg argument");

}

Result:

ERROR. constructor Animal () not found

So what happened?. We have said before that if at least one constructor is defined, Java doesn't provide one by default. The error occurs because Reptile(int l) will call Animal() before the body of the constructor is executed as if a super() was there. We don't need to get back to our Animal() constructor. We only need to customize super() which is currently implicit:

CONSTRUCTORS AND SUBCLASSING

2007-12-01T12:22:00.000+05:30

Constructors are not inherited into subclasses; you must define each form of constructor that you require. A class constructor that has no constructors defined in the source is given exactly one constructor. This is the default constructor; it takes no arguments and is of public accessibility
Java insists that the object is initialized from the top of the class hierarchy downward:
Example:
class Test {
static int i = 0;
public static void main(String args[]) {
Terrier t = new Terrier(15);
System.out.println("Count: "+ i);
}
}
class Animal {
Animal (int increment) {
Test.i += increment;
System.out.println("Animal constructor");
}
}
class Dog extends Animal {
Dog (int increment) {
super(increment);
Test.i += increment;
System.out.println("Dog constructor");
}
}
class Terrier extends Dog {
Terrier (int increment) {
super(increment);
Test.i += increment;
System.out.println("Terrier constructor");
}
}
Result:
Animal constructor
Dog constructor
Terrier constructor
Count: 45

Calling Superclass Constructors
In the case of constructors you also use the keyword super() as with plain methods but you don't actually need to supply the name of the constructor, just super(arguments).
Example:
class Test {
public static void main(String args[]) {
Dog d = new Dog();
Animal a = new Animal(); // ERROR. No constructor
System.out.println("Legs: "+ d.legs);
System.out.println("Mammal: "+ d.mammal);
System.out.println("Vertebrate: "+ d.vertebrate);
}
}
class Animal {
int legs = 0;
boolean mammal = false;
boolean vertebrate = false;
Animal (int l, boolean m, boolean v) {
legs = l;
mammal = m;
vertebrate = v;
System.out.println("Animal constructor");
}
}
class Dog extends Animal {
Dog () {
super(4,true,true);
System.out.println("Dog constructor");
}
}
Result:
Animal constructor
Dog constructor
Legs: 4
Mammal: true
Vertebrate: true
As you can see, if you provide a valid constructor, Java doesn't generate a default constructor anymore. To make the erroneous line valid, you should add the constructor to the Animal class:
Animal() {}
That also means that if you define a argument-based constructor in the superclass, you won't be able to create non-argument objects out of any of the subclasses. For example, suppose you want to add a Bird class to the previous example:
class Bird extends Animal {
Bird() {
System.out.println("Bird constructor");
}
}
You have to define Animal() {} in the Animal if you want to use this new class like this:
Bird b = new Bird();
This happens because Java will try to call Animal() before calling Bird(), and Animal() hasn't been defined. You can specify which constructor of the superclass will be called by using super:
class Bird extends Animal {
Bird() {
super(2,false,true);
System.out.println("Bird constructor");
}
}
Please note that super() must be the first statement in the Bird() constructor in order to be recognized by the compiler.

OVERRIDING

2007-12-01T12:20:00.000+05:30

When you extend a class to produce a new one, you inherit and have access to all the non-private methods of the original class. When you want to redefine the behavior of one of these methods to suit your new class, you override the method to define a new one.
An overriding method replaces the method it overrides. Each method in a parent class can be overridden at most once in any subclass. Overriding methods should have argument lists of identical type and order. The return type of an overriding method must be identical to that of the method it overrides.
Some other special rules apply to overriding methods:
• The accessibility must not be more restricted than the original method
• The method must not throw checked exceptions of classes that are not possible for the original method.
Example #1:
class Animal {
void saySomething() {
System.out.println("Our kind is unable to speak");
}
}
class Dog extends Animal {
void saySomething() {
System.out.println("Wof!");
}
}
Animal a[] = new Animal[2];
a[0] = new Animal();
a[1] = new Dog();
a[0].saySomething();
a[1].saySomething();
Result:
Our kind is unable to speak
Wof!
Invoking Overridden Methods
It is very useful to be able to invoke an overridden method from the method that overrides it. The keyword “super” allows to access features of a class "up" in the hierarchy.
Example:
class Animal {
void saySomething() {
System.out.println("Our kind is unable to speak");
}
}
class Dog extends Animal {
void saySomething() {
super.saySomething();
System.out.println("Wof!");
}
}
Dog d = new Dog();
d.saySomething();
Result:
Our kind is unable to speak
Wof!

OVERLOADING

2007-12-01T12:19:00.000+05:30

Overloading is about reusing the same method name with different arguments and sometimes different return type. Some basic rules apply:
• In the class that defines de original method, or a subclass of that class, the method name can be reused if the argument list differs in terms of the type of at least one argument.
• A difference in return type alone is insufficient to constitute an overload and is illegal.
Example #1:
public static void main(String args[]) {
System.out.println( DoubleIt("Hello world") );
System.out.println( DoubleIt(15) );
}
static String DoubleIt(String name) {
name += " " + name;
return name;
}
static String DoubleIt(int id) {
id *= 2;
return Integer.toString(id);
}
Result:
Hello Hello
30
Example #2:
int MyMethod(int a) { }
float MyMethod(int a) { } // ERROR!!!. Difference in return type alone
// is insufficient to constitute an overload.
Example #3:
int MyMethod(int value) { }
float MyMethod(int number) { } // ERROR!!!. The types should be
// different, not the variable names.
Invoking Overloaded Methods
Method overloaded names are effectively independent methods. Using the same name is just a convenience to the programmer. Overloaded methods may call one another simply by providing a normal method call with an appropriately formed argument list.
Example:
public static void main(String args[]) {
allStars(15);
}
static void allStars(String name) {
name = "*** " + name + " ***";
System.out.println(name);
}
static void allStars(int id) {
String s = ""+id;
allStars(s);
}
In this example, we have created a method to show names surrounded by stars, we also wanted to accept int values, so we added a method that receives ints but converts them to strings and passes the result to the original string method.

JAVA DUMPS 12(SCJP)

2007-11-29T12:36:00.001+05:30

QUESTION NO: 60

Given:

1. public class Foo {

2. public void main( String[] args ) {

3. System.out.printIn( "Hello" + args[0] );

4. }

5. }

What is the result if this code is executed with the command line?

java Foo world

A. Hello

B. Hello Foo

C. Hello world

D. Compilation fails.

E. The code does not run.

Answer: E

QUESTION NO: 61

Given:

11. public void foo( boolean a, boolean b ){

12. if( a ) {

13. System.out.println( "A" );

14. } else if ( a && b ) {

15. System.out.println( "A&&B" );

16. } else {

17. if ( !b ) {

18. System.out.println( "notB" );

19. } else {

20. System.out.println( "ELSE" );

21. }

22. }

23. }

What is correct?

A. If a is true and b is true then the output is "A&&B".

B. If a is true and b is false then the output is "notB".

C. If a is false and b is true then the output is "ELSE".

D. If a is false and b is false then the output is "ELSE".

Answer: C

QUESTION NO: 62

Which two cause a compiler error? (Choose two)

A. int[] scores = {3, 5, 7};

B. int [][] scores = {2,7,6}, {9,3,45};

C. String cats[] = {"Fluffy", "Spot", "Zeus"};

D. boolean results[] = new boolean [3] {true, false, true};

E. Integer results[] = {new Integer(3), new Integer(5), new

Integer(8)};

F. String[] dogs = new String[]{new String("Fido"),new

String("Spike"), new String("Aiko")};

Answer: B, D

QUESTION NO: 63

Given:

11. int i = 0, j = 5;12. tp; for (;;) {

12. i++;

13. for(;;) {

14. if (i> --j) {

15. break tp;

16. break tp;

17. }

18. }

19. System.out.printIn("i=" +i ",j ="+j);

What is the result?

A. i = 1, j = 0

B. i = 1, j = 4

C. i = 3, j = 4

D. i = 3, j = 0

E. Compilation fails.

Answer: E

QUESTION NO: 64

Given:

1. public abstract class Test {

2. public abstract void methodA();

4. public abstract void methodB()

5. {

6. System.out.println("Hello");

7. }

8. }

Which two changes, independently applied, allow this code to compile? (Choose two)

A. Add a method body to methodA.

B. Replace lines 5 – 7 with a semicolon (";").

C. Remove the abstract qualifier from the declaration of Test.

D. Remove the abstract qualifier from the declaration of methodA.

E. Remove the abstract qualifier from the declaration of methodB.

Answer: B, E

QUESTION NO: 65

Given:

1. public class Test {

2. public static void main(String Args[]) {

3. int i =1, j = 0;

4. switch(i) {

5. case 2: j +=6;

6. case 4: j +=1;

7. default: j +=2;

8. case 0: j +=4;

9. }

10. System.out.printIn("j =" +j);

11. }

12. }

What is the result?

A. 0

B. 2

C. 4

D. 6

E. 9

F. 13

Answer: D

QUESTION NO: 66

Given:

1. class A {

2. }

3. class Alpha {

4. private A myA = new A();

6. void dolt( A a ) {

7. a = null;

8. }

9. void tryIt() {

10. dolt( myA );

11. }

12. }

Which two statements are correct? (Choose two)

A. There are no instanced of A that will become eligible for garbage collection.

B. Explicitly setting myA to null marks that instance to be eligible for garbage collection.

C. Any call on tryIt() causes the private instance of A to be marked for garbage

collection.

D. Private instances of A become eligible for garbage collection when instances of Alpha

become eligible for garbage collection.

Answer: B, D