Pages

Showing posts with label Certification. Show all posts
Showing posts with label Certification. Show all posts

Aug 25, 2012

Exercise-7 (Collections and Generics)



1. Given: 
public static void main(String[] args) {
// INSERT DECLARATION HERE
for (int i = 0; i <= 10; i++) {
List<Integer> row = new ArrayList<Integer>();
for (int j = 0; j <= 10; j++)
row.add(i * j);
table.add(row);
}
for (List<Integer> row : table)
System.out.println(row);
}
Which statements could be inserted at // INSERT DECLARATION HERE to allow this code to compile and run? (Choose all that apply.)
A. List<List<Integer>> table = new List<List<Integer>>();
B. List<List<Integer>> table = new ArrayList<List<Integer>>();
C. List<List<Integer>> table = new ArrayList<ArrayList<Integer>>();
D. List<List, Integer> table = new List<List, Integer>();
E. List<List, Integer> table = new ArrayList<List, Integer>();
F. List<List, Integer> table = new ArrayList<ArrayList, Integer>();
G. None of the above
Answer:
B is correct.
A is incorrect because List is an interface, so you can't say new List() regardless of any generic types. D, E, and F are incorrect because List only takes one type parameter (a Map would take two, not a List). C is tempting, but incorrect. The type argument <List<Integer>> must be the same for both sides of the assignment, even though the constructor new ArrayList() on the right side is a subtype of the declared type List on the left. (Objective 6.4)

2. Which statements are true about comparing two instances of the same class, given that the equals() and hashCode() methods have been properly overridden? (Choose all that apply.)
A. If the equals() method returns true, the hashCode() comparison == might return false
B. If the equals() method returns false, the hashCode() comparison == might return true
C. If the hashCode() comparison == returns true, the equals() method must return true
D. If the hashCode() comparison == returns true, the equals() method might return true
E. If the hashCode() comparison != returns true, the equals() method might return true
Answer:
B and D. B is true because often two dissimilar objects can return the same hashcode value. D is true because if the hashCode() comparison returns ==, the two objects might or might not be equal.
A, C, and E are incorrect. C is incorrect because the hashCode() method is very flexible in its return values, and often two dissimilar objects can return the same hash code value. A and E are a negation of the hashCode() and equals() contract. (Objective 6.2)

3. Given: 
public static void before() {
Set set = new TreeSet();
set.add("2");
set.add(3);
set.add("1");
Iterator it = set.iterator();
while (it.hasNext())
System.out.print(it.next() + " ");
}
Which statements are true?
A. The before() method will print 1 2
B. The before() method will print 1 2 3
C. The before() method will print three numbers, but the order cannot be determined
D. The before() method will not compile
E. The before() method will throw an exception at runtime
Answer:
E is correct. You can't put both Strings and ints into the same TreeSet. Without generics, the compiler has no way of knowing what type is appropriate for this TreeSet, so it allows everything to compile. At runtime, the TreeSet will try to sort the elements as they're added, and when it tries to compare an Integer with a String it will throw a ClassCastException. Note that although the before() method does not use generics, it does use autoboxing. Watch out for code that uses some new features and some old features mixed together.
A, B, C, and D are incorrect based on the above. (Objective 6.5)

4. Given: 
import java.util.*;
class MapEQ {
public static void main(String[] args) {
Map<ToDos, String> m = new HashMap<ToDos, String>();
ToDos t1 = new ToDos("Monday");
ToDos t2 = new ToDos("Monday");
ToDos t3 = new ToDos("Tuesday");
m.put(t1, "doLaundry");
m.put(t2, "payBills");
m.put(t3, "cleanAttic");
System.out.println(m.size());
} }
class ToDos{
String day;
ToDos(String d) { day = d; }
public boolean equals(Object o) {
return ((ToDos)o).day == this.day;
}
// public int hashCode() { return 9; }
}
Which is correct? (Choose all that apply.)
A. As the code stands it will not compile
B. As the code stands the output will be 2
C. As the code stands the output will be 3
D. If the hashCode() method is uncommented the output will be 2
E. If the hashCode() method is uncommented the output will be 3
F. If the hashCode() method is uncommented the code will not compile
Answer:
C and D are correct. If hashCode() is not overridden then every entry will go into its own bucket, and the overridden equals() method will have no effect on determining equivalency. If hashCode() is overridden, then the overridden equals() method will view t1 and t2 as duplicates.
A, B, E, and F are incorrect based on the above. (Objective 6.2)

5. Given: 
public class AccountManager {
private Map accountTotals = new HashMap(); //line 13
private int retirementFund;

public int getBalance(String accountName) {
Integer total = (Integer) accountTotals.get(accountName);  // line 17
if (total == null)
total = Integer.valueOf(0);
return total.intValue();   // line 20
}
public void setBalance(String accountName, int amount) {
accountTotals.put(accountName, Integer.valueOf(amount)); // line 24
} }
This class is to be updated to make use of appropriate generic types, with no changes in behavior (for better or worse). Which of these steps could be performed? (Choose three.)
A. Replace line 13 with private
Map<String, int> accountTotals = new HashMap<String, int>();
B. Replace line 13 with private
Map<String, Integer> accountTotals = new HashMap<String, Integer>();
C. Replace line 13 with
private Map<String<Integer>> accountTotals = new HashMap<String<Integer>>();
D. Replace lines 17-20 with
int total = accountTotals.get(accountName);
if (total == null) total = 0;
return total;
E. Replace lines 17-20 with
Integer total = accountTotals.get(accountName);
if (total == null) total = 0; return total;
F. Replace lines 17-20 with return accountTotals.get(accountName);
G. Replace line 24 with accountTotals.put(accountName, amount);
H. Replace line 24 with accountTotals.put(accountName, amount.intValue());
Answer:
B, E, and G are correct.
A is wrong because you can't use a primitive type as a type parameter. C is wrong because a Map takes two type parameters separated by a comma. D is wrong because an int can't autobox to a null, and F is wrong because a null can't unbox to 0. H is wrong because you can't autobox a primitive just by trying to invoke a method with it. (Objective 6.4)

6. Given: 
interface Hungry<E> { void munch(E x); }
interface Carnivore<E extends Animal> extends Hungry<E> {}
interface Herbivore<E extends Plant> extends Hungry<E> {}
abstract class Plant {}
class Grass extends Plant {}
abstract class Animal {}
class Sheep extends Animal implements Herbivore<Sheep> {
public void munch(Sheep x) {}
}
class Wolf extends Animal implements Carnivore<Sheep> {
public void munch(Sheep x) {}
}
Which of the following changes (taken separately) would allow this code to compile? (Choose all that apply.)
A. Change the Carnivore interface to
interface Carnivore<E extends Plant> extends Hungry<E> {}
B. Change the Herbivore interface to
interface Herbivore<E extends Animal> extends Hungry<E> {}
C. Change the Sheep class to
class Sheep extends Animal implements Herbivore<Plant> {
public void munch(Grass x) {}
}
D. Change the Sheep class to
class Sheep extends Plant implements Carnivore<Wolf> {
public void munch(Wolf x) {}
}
E. Change the Wolf class to
class Wolf extends Animal implements Herbivore<Grass> {
public void munch(Grass x) {}
}
F. No changes are necessary

Answer:
B is correct. The problem with the original code is that Sheep tries to implement HerbivoresSheep> and Herbivore declares that its type parameter E can be any type that extends Plant. Since a Sheep is not a Plant, Herbivore<Sheep> makes no sense- the type Sheep is outside the allowed range of Herbivore's parameter E. Only solutions that either alter the definition of a Sheep or alter the definition of Herbivore will be able to fix this. So A, E, and F are eliminated. B works, changing the definition of an Herbivore to allow it to eat Sheep solves the problem. C doesn't work because an Herbivore<Plant> must have a munch(Plant) method, not munch(Grass). And D doesn't work, because in D we made Sheep extend Plant, now the Wolf class breaks because its munch(Sheep) method no longer fulfills the contract of Carnivore. (Objective 6.4)

7. Which collection class(es) allows you to grow or shrink its size and provides indexed access to its elements, but whose methods are not synchronized? (Choose all that apply.)
A. java.util.HashSet
B. java.util.LinkedHashSet
C. java.util.List
D. java.util.ArrayList
E. java.util.Vector
F. java.util.PriorityQueue
Answer:
D is correct. All of the collection classes allow you to grow or shrink the size of your collection. ArrayList provides an index to its elements. The newer collection classes tend not to have synchronized methods. Vector is an older implementation of ArrayList functionality and has synchronized methods; it is slower than ArrayList.
A, B, C, E, and F are incorrect based on the logic described above; Notes: C, List is an interface, and F, PriorityQueue does not offer access by index. (Objective 6.1)

8. Given a method declared as 
public static <E extends Number> List<E> process(List<E> nums)
A programmer wants to use this method like this 
// INSERT DECLARATIONS HERE
output = process(input);
Which pairs of declarations could be placed at // INSERT DECLARATIONS HERE to allow the code to compile? (Choose all that apply.)
A. ArrayList<Integer> input = null; ArrayList<Integer> output = null;
B. ArrayList<Integer> input = null; List<Integer> output = null;
C. ArrayList<Integer> input = null; List<Number> output = null;
D. List<Number> input = null; ArrayList<Integer> output = null;
E. List<Number> input = null; List<Number> output = null;
F. List<Integer> input = null; List<Integer> output = null;
G. None of the above
Answer:
B, E, and F are correct.
The return type of process is definitely declared as a List, not an ArrayList, so A and D are wrong. C is wrong because the return type evaluates to List<Integer>, and that can't be assigned to a variable of type List<Number>. Of course all these would probably cause a NullPointerException since the variables are still null-but the question only asked us to get the code to compile. (Objective 6.4)

9. Given the proper import statement(s), and 
PriorityQueue<String> pq = new PriorityQueue<String>();
pq.add("2");
pq.add("4");
System.out.print(pq.peek() + " ");
pq.offer("1");
pq.add("3");
pq.remove("1");
System.out.print(pq.poll() + " ");
if(pq.remove("2")) System.out.print(pq.poll() + " ");
System.out.println(pq.poll() + " " + pq.peek());
What is the result?
A. 2 2 3 3
B. 2 2 3 4
C. 4 3 3 4
D. 2 2 3 3 3
E. 4 3 3 3 3
F. 2 2 3 3 4
G. Compilation fails
H. An exception is thrown at runtime
Answer:
B is correct. For the sake of the exam, add() and offer() both add to (in this case), naturally sorted queues. The calls to poll() both return and then remove the first item from the queue, so the if test fails.
A, C, D, E, F, G, and H are incorrect based on the above. (Objective 6.1)

10. Given: 
import java.util.*;
public class Mixup {
public static void main(String[] args) {
Object o = new Object();
// insert code here   line 7
s.add("o");
s.add(o);
}
}
And these three fragments:
I. Set s = new HashSet();
II. TreeSet s = new TreeSet();
III. LinkedHashSet s = new LinkedHashSet();
When fragments I, II, or III are inserted, independently, at line 7, which are true? (Choose all that apply.)
A. Fragment I compiles
B. Fragment II compiles
C. Fragment III compiles
D. Fragment I executes without exception
E. Fragment II executes without exception
F. Fragment III executes without exception
Answer:
A, B, C, D, and F are all correct.
Only E is incorrect. Elements of a TreeSet must in some way implement Comparable. (Objective 6.1)

11. Given: 
import java.util.*;
class Turtle {
int size;
public Turtle(int s) { size = s; }
public boolean equals(Object o) { return (this.size == ((Turtle)o).size); }
// insert code here line 8
}
public class TurtleTest {
public static void main(String[] args) {
LinkedHashSet<Turtle> t = new LinkedHashSet<Turtle>();
t.add(new Turtle(1)); t.add(new Turtle(2)); t.add(new Turtle(1));
System.out.println(t.size());
}
}
And these two fragments:
I. public int hashCode() { return size/5; }
II. // no hashCode method declared
If fragment I or II is inserted, independently, at line 8, which are true? (Choose all that apply.)
A. If fragment I is inserted, the output is 2
B. If fragment I is inserted, the output is 3
C. If fragment II is inserted, the output is 2
D. If fragment II is inserted, the output is 3
E. If fragment I is inserted, compilation fails
F. If fragment II is inserted, compilation fails
Answer:
A and D are correct. While fragment II wouldn't fulfill the hashCode() contract (as you can see by the results), it is legal Java. For the purpose of the exam, if you don't override hashCode(), every object will have a unique hashcode.
B, C, E, and F are incorrect based on the above. (Objective 6.2)

12. Given the proper import statement(s), and: 
TreeSet<String> s = new TreeSet<String>();
 TreeSet<String> subs = new TreeSet<String>();
s.add("a"); s.add("b"); s.add("c"); s.add("d"); s.add("e");

subs = (TreeSet)s.subSet("b", true, "d", true);
s.add("g");
s.pollFirst();
s.pollFirst();
s.add("c2");
System.out.println(s.size() +" "+ subs.size());
Which are true? (Choose all that apply.)
A. The size of s is 4
B. The size of s is 5
C. The size of s is 7
D. The size of subs is 1
E. The size of subs is 2
F. The size of subs is 3
G. The size of subs is 4
H. An exception is thrown at runtime
Answer:
B and F are correct. After "g" is added, TreeSet s contains six elements and TreeSet subs contains three (b, c, d), because "g" is out of the range of subs. The first pollFirst() finds and removes only the "a". The second pollFirst() finds and removes the "b" from both TreeSets (remember they are backed). The final add() is in range of both TreeSets. The final contents are [c,c2,d,e,g] and [c,c2,d].
A, C, D, E, G, and H are incorrect based on the above. (Objective 6.3)

13. Given: 
import java.util.*;
public class Magellan {
public static void main(String[] args) {
TreeMap<String, String> myMap = new TreeMap<String, String>();
myMap.put("a", "apple"); myMap.put("d", "date");
myMap.put("f", "fig"); myMap.put("p", "pear");
System.out.println("1st after mango: " + // sop 1
myMap.higherKey("f"));
System.out.println("1st after mango: " + // sop 2
myMap.ceilingKey("f"));
System.out.println("1st after mango: " + // sop 3
myMap.floorKey("f"));
SortedMap<String, String> sub = new TreeMap<String, String>();
sub = myMap.tailMap("f");
System.out.println("1st after mango: " + // sop 4
sub.firstKey());
}
}
Which of the System.out.println statements will produce the output 1st after mango: p? (Choose all that apply.)
A. sop 1
B. sop 2
C. sop 3
D. sop 4
E. None; compilation fails
F. None; an exception is thrown at runtime
Answer:
A is correct. The ceilingKey() method's argument is inclusive. The floorKey() method would be used to find keys before the specified key. The firstKey() method's argument is also inclusive.
B, C, D, E, and F are incorrect based on the above. (Objective 6.3)

14. Given: 
import java.util.*;
class Business { }
class Hotel extends Business { }
class Inn extends Hotel { }
public class Travel {
ArrayList<Hotel> go() {
// insert code here  line 9
}
}
Which, inserted independently at line 9, will compile? (Choose all that apply.)
A. return new ArrayList<Inn>();
B. return new ArrayList<Hotel>();
C. return new ArrayList<Object>();
D. return new ArrayList<Business>();
Answer:
B is correct.
A is incorrect because polymorphic assignments don't apply to generic type parameters. C and D are incorrect because they don't follow basic polymorphism rules. (Objective 6.4)

15. Given: 
import java.util.*;
class Dog { int size; Dog(int s) { size = s; } }
public class FirstGrade {
public static void main(String[] args) {
TreeSet<Integer> i = new TreeSet<Integer>();
TreeSet<Dog> d = new TreeSet<Dog>();
d.add(new Dog(1)); d.add(new Dog(2)); d.add(new Dog(1));
i.add(1); i.add(2); i.add(1);
System.out.println(d.size() + " " + i.size());
}
}
What is the result?
A. 1 2
B. 2 2
C. 2 3
D. 3 2
E. 3 3
F. Compilation fails
G. An exception is thrown at runtime
Answer:
G is correct. Class Dog needs to implement Comparable in order for a TreeSet (which keeps its elements sorted) to be able to contain Dog objects.
A, B, C, D, E, and F are incorrect based on the above. (Objective 6.5)

16. Given: 
import java.util.*;
public class GeoCache {
public static void main(String[] args) {
String[] s = {"map", "pen", "marble", "key"};
Othello o = new Othello();
Arrays.sort(s,o);
for(String s2: s) System.out.print(s2 + " ");
System.out.println(Arrays.binarySearch(s, "map"));
}
static class Othello implements Comparator<String> {
public int compare(String a, String b) { return b.compareTo(a); }
}
}
Which are true? (Choose all that apply.)
A. Compilation fails
B. The output will contain a 1
C. The output will contain a 2
D. The output will contain a -1
E. An exception is thrown at runtime
F. The output will contain "key map marble pen"
G. The output will contain "pen marble map key"
Answer:
D and G are correct. First, the compareTo() method will reverse the normal sort. Second, the sort() is valid. Third, the binarySearch() gives -1 because it needs to be invoked using the same Comparator (o), as was used to sort the array. Note that when the binarySearch() returns an "undefined result" it doesn't officially have to be a -1, but it usually is, so if you selected only G, you get full credit!
A, B, C, E, and F are incorrect based on the above. (Objective 6.5)


regards,
TechDexters Support Team
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses
Posted by 0 comments

Aug 17, 2012

OCJP Exercise -6 (Strings)



1. Given: 
import java.util.regex.*;
class Regex2 {
public static void main(String[] args) {
Pattern p = Pattern.compile(args[0]);
Matcher m = p.matcher(args[1]);
boolean b = false;
while(b = m.find()) {
System.out.print(m.start() + m.group());
}
}
}
And the command line:
java Regex2 "\d*" ab34ef
What is the result?
A. 234
B. 334
C. 2334
D. 0123456
E. 01234456
F. 12334567
G. Compilation fails
Answer:
E is correct. The \d is looking for digits. The * is a quantifier that looks for 0 to many occurrences of the pattern that precedes it. Because we specified *, the group() method returns empty Strings until consecutive digits are found, so the only time group() returns a value is when it returns 34 when the matcher finds digits starting in position 2. The start() method returns the starting position of the previous match because, again, we said find 0 to many occurrences.
A, B, C, D, F, and G are incorrect based on the above. (Objective 3.5)

2. Given: 
import java.io.*;
class Player {
Player() { System.out.print("p"); }
}
class CardPlayer extends Player implements Serializable {
CardPlayer() { System.out.print("c"); }
public static void main(String[] args) {
CardPlayer c1 = new CardPlayer();
try {
FileOutputStream fos = new FileOutputStream("play.txt");
ObjectOutputStream os = new ObjectOutputStream(fos);
os.writeObject(c1);
os.close();
FileInputStream fis = new FileInputStream("play.txt");
ObjectInputStream is = new ObjectInputStream(fis);
CardPlayer c2 = (CardPlayer) is.readObject();
is.close();
} catch (Exception x ) { }
}
}
What is the result?
A. pc
B. pcc
C. pcp
D. pcpc
E. Compilation fails
F. An exception is thrown at runtime
Answer:
C is correct. It's okay for a class to implement Serializable even if its superclass doesn't. However, when you deserialize such an object, the non-serializable superclass must run its constructor. Remember, constructors don't run on deserialized classes that implement Serializable.
A, B, D, E, and F are incorrect based on the above. (Objective 3.3)

3. Given: 
class TKO {
public static void main(String[] args) {
String s = "-";
Integer x = 343;
long L343 = 343L;
if(x.equals(L343)) s += ".e1 ";
if(x.equals(343)) s += ".e2 ";
Short s1 = (short)((new Short((short)343)) / (new Short((short)49)));
if(s1 == 7) s += "=s ";
if(s1 < new Integer(7+1)) s += "fly ";
System.out.println(s);
} }
Which of the following will be included in the output String s? (Choose all that apply.)
A. .e1
B. .e2
C. =s
D. fly
E. None of the above
F. Compilation fails
G. An exception is thrown at runtime
Answer:
B, C, and D are correct. Remember, that the equals() method for the integer wrappers will only return true if the two primitive types and the two values are equal. With C, it's okay to unbox and use ==. For D, it's okay to create a wrapper object with an expression, and unbox it for comparison with a primitive.
A, E, F, and G are incorrect based on the above. (Remember that A is using the equals() method to try to compare two different types.) (Objective 3.1)

4. Given: 
import java.io.*;
class Keyboard { }
public class Computer implements Serializable {
	
		private Keyboard k = new Keyboard();
public static void main(String[] args) {
Computer c = new Computer();
c.storeIt(c);
}
void storeIt(Computer c) {
try {
ObjectOutputStream os = new ObjectOutputStream(
new FileOutputStream("myFile"));
os.writeObject(c);
os.close();
System.out.println("done");
} catch (Exception x) {System.out.println("exc"); }
}
}
What is the result? (Choose all that apply.)
A. exc
B. done
C. Compilation fails
D. Exactly one object is serialized
E. Exactly two objects are serialized
Answer:
A is correct. An instance of type Computer Has-a Keyboard. Because Keyboard doesn't implement Serializable, any attempt to serialize an instance of Computer will cause an exception to be thrown.
B, C, D, and E are incorrect based on the above. If Keyboard did implement Serializable then two objects would have been serialized. (Objective 3.3)

5. Using the fewest fragments possible (and filling the fewest slots possible), complete the code below so that the class builds a directory named "dir3" and creates a file named "file3" inside "dir3". Note you can use each fragment either zero or one times. Code: 
import java.io.______________
class Maker {
public static void main(String[] args) {
___________ ___________ ___________
___________ ___________ ___________
___________ ___________ ___________
___________ ___________ ___________
___________ ___________ ___________
___________ ___________ ___________
___________ ___________ ___________
} }
Fragments:
File; FileDescriptor; FileWriter; Directory;
try { .createNewDir(); File dir File
{ } (Exception x) ("dir3"); file
file .createNewFile(); = new File = new File
dir (dir, "file3"); (dir, file); .createFile();
} catch ("dir3", "file3"); .mkdir(); File file

Answer: 
import java.io.File;
class Maker {
public static void main(String[] args) {
try {
File dir = new File("dir3");
dir.mkdir();
File file = new File(dir, "file3");
file.createNewFile();
} catch (Exception x) { }
} }

Notes: The new File statements don't make actual files or directories, just objects. You need the mkdir() and createNewFile() methods to actually create the directory and the file. (Objective 3.2)

6. Given that 1119280000000L is roughly the number of milliseconds from Jan. 1, 1970, to June 20, 2005, and that you want to print that date in German, using the LONG style such that "June" will be displayed as "Juni", complete the code using the fragments below. Note: you can use each fragment either zero or one times, and you might not need to fill all of the slots. Code: 
import java.___________
import java.___________
class DateTwo {
public static void main(String[] args) {
Date d = new Date(1119280000000L);
DateFormat df = ___________________________
________________ , _________________ );
System.out.println(________________
}
}
Fragments:
io.*; new DateFormat( Locale.LONG
nio.*; DateFormat.getInstance( Locale.GERMANY
util.*; DateFormat.getDateInstance( DateFormat.LONG
text.*; util.regex; DateFormat.GERMANY
date.*; df.format(d)); d.format(df));

Answer:
import java.util.*; import java.text.*; class DateTwo { public static void main(String[] args) { Date d = new Date(1119280000000L); DateFormat df = DateFormat.getDateInstance( DateFormat.LONG, Locale.GERMANY); System.out.println(df.format(d)); } } Notes: Remember that you must build DateFormat objects using static methods. Also remember that you must specify a Locale for a DateFormat object at the time of instantiation. The getInstance() method does not take a Locale. (Objective 3.4)

7. Given: 
import java.io.*;
class Directories {
static String [] dirs = {"dir1", "dir2"};
public static void main(String [] args) {
for (String d : dirs) {
// insert code 1 here
File file = new File(path, args[0]);
// insert code 2 here
}
}
}
and that the invocation
java Directories file2.txt
is issued from a directory that has two subdirectories, "dir1" and "dir2", and that "dir1" has a file "file1.txt" and "dir2" has a file "file2.txt", and the output is "false true", which set(s) of code fragments must be inserted? (Choose all that apply.)
A. String path = d; System.out.print(file.exists() + " ");
B. String path = d; System.out.print(file.isFile() + " ");
C. String path = File.separator + d; System.out.print(file.exists() + " ");
D. String path = File.separator + d; System.out.print(file.isFile() + " ");
Answer:
A and B are correct. Because you are invoking the program from the directory whose direct subdirectories are to be searched, you don't start your path with a File.separator character. The exists() method tests for either files or directories; the isFile() method tests only for files. Since we're looking for a file, both methods work.
C and D are incorrect based on the above. (Objective 3.2)

8. Given: 
import java.io.*;
public class TestSer {
public static void main(String[] args) {
SpecialSerial s = new SpecialSerial();
try {
ObjectOutputStream os = new ObjectOutputStream(
new FileOutputStream("myFile"));
os.writeObject(s); os.close();
System.out.print(++s.z + " ");
ObjectInputStream is = new ObjectInputStream(
new FileInputStream("myFile"));
SpecialSerial s2 = (SpecialSerial)is.readObject();
is.close();
System.out.println(s2.y + " " + s2.z);
} catch (Exception x) {System.out.println("exc"); }
}
}
class SpecialSerial implements Serializable {
transient int y = 7;
static int z = 9;
}
Which are true? (Choose all that apply.)
A. Compilation fails
B. The output is 10 0 9
C. The output is 10 0 10
D. The output is 10 7 9
E. The output is 10 7 10
F. In order to alter the standard deserialization process you would implement the readObject() method in SpecialSerial
G. In order to alter the standard deserialization process you would implement the defaultReadObject() method in SpecialSerial
Answer:
C and F are correct. C is correct because static and transient variables are not serialized when an object is serialized. F is a valid statement.
A, B, D, and E are incorrect based on the above. G is incorrect because you don't implement the defaultReadObject() method, you call it from within the readObject()method, along with any custom read operations your class needs. (Objective 3.3)

9. Given: 
3. public class Theory {
4. public static void main(String[] args) {
5. String s1 = "abc";
6. String s2 = s1;
7. s1 += "d";
8. System.out.println(s1 + " " + s2 + " " + (s1==s2));
9.
10. StringBuffer sb1 = new StringBuffer("abc");
11. StringBuffer sb2 = sb1;
12. sb1.append("d");
13. System.out.println(sb1 + " " + sb2 + " " + (sb1==sb2));
14. }
15. }
Which are true? (Choose all that apply.)
A. Compilation fails
B. The first line of output is abc abc true
C. The first line of output is abc abc false
D. The first line of output is abcd abc false
E. The second line of output is abcd abc false
F. The second line of output is abcd abcd true
G. The second line of output is abcd abcd false
Answer:
D and F are correct. While String objects are immutable, references to Strings are mutable. The code s1 += "d"; creates a new String object. StringBuffer objects are mutable, so the append() is changing the single StringBuffer object to which both StringBuffer references refer.
A, B, C, E, and G are incorrect based on the above. (Objective 3.1)

10. Given: 
3. import java.io.*;
4. public class ReadingFor {
5. public static void main(String[] args) {
6. String s;
7. try {
8. FileReader fr = new FileReader("myfile.txt");
9. BufferedReader br = new BufferedReader(fr);
10. while((s = br.readLine()) != null)
11. System.out.println(s);
12. br.flush();
13. } catch (IOException e) { System.out.println("io error"); }
16. }
17. }
And given that myfile.txt contains the following two lines of data:
ab cd
What is the result?
A. ab
B. abcd
C. ab cd
D. a b c d
E. Compilation fails
Answer:
E is correct. You need to call flush() only when you're writing data. Readers don't have flush() methods. If not for the call to flush(), answer C would be correct.
A, B, C, and D are incorrect based on the above. (Objective 3.2)

11. Given: 
3. import java.io.*;
4. public class Talker {
5. public static void main(String[] args) {
6. Console c = System.console();
7. String u = c.readLine("%s", "username: ");
8. System.out.println("hello " + u);
9. String pw;
10. if(c != null && (pw = c.readPassword("%s", "password: ")) != null)
11. // check for valid password
12. }
13. }
If line 6 creates a valid Console object, and if the user enters fred as a username and 1234 as a password, what is the result? (Choose all that apply.)
A.
username:
password:
B.
username: fred
password:
C.
username: fred
password: 1234
D. Compilation fails
E. An exception is thrown at runtime
Answer:
D is correct. The readPassword() method returns a char[]. If a char[] were used, answer B would be correct.
A, B, C, and E are incorrect based on the above. (Objective 3.2)

12. Given: 
3. import java.io.*;
4. class Vehicle { }
5. class Wheels { }
6. class Car extends Vehicle implements Serializable { }
7. class Ford extends Car { }
8. class Dodge extends Car {
9. Wheels w = new Wheels();
10. }
Instances of which class(es) can be serialized? (Choose all that apply.)
A. Car
B. Ford
C. Dodge
D. Wheels
E. Vehicle
Answer:
A and B are correct. Dodge instances cannot be serialized because they "have" an instance of Wheels, which is not serializable. Vehicle instances cannot be serialized even though the subclass Car can be.
C, D, and E are incorrect based on the above. (Objective 3.3)

13. Given: 
3. import java.text.*;
4. public class Slice {
5. public static void main(String[] args) {
6. String s = "987.123456";
7. double d = 987.123456d;
8. NumberFormat nf = NumberFormat.getInstance();
9. nf.setMaximumFractionDigits(5);
10. System.out.println(nf.format(d) + " ");
11. try {
12. System.out.println(nf.parse(s));
13. } catch (Exception e) { System.out.println("got exc"); }
14. }
15. }
Which are true? (Choose all that apply.)
A. The output is 987.12345 987.12345
B. The output is 987.12346 987.12345
C. The output is 987.12345 987.123456
D. The output is 987.12346 987.123456
E. The try/catch block is unnecessary
F. The code compiles and runs without exception
G. The invocation of parse() must be placed within a try/catch block
Answer:
D, F, and G are correct. The setMaximumFractionDigits() applies to the formatting but not the parsing. The try/catch block is placed appropriately. This one might scare you into thinking that you'll need to memorize more than you really do. If you can remember that you're formatting the number and parsing the string you should be fine for the exam.
A, B, C, and E are incorrect based on the above. (Objective 3.4)

14. Given: 
3. import java.util.regex.*;
4. public class Archie {
5. public static void main(String[] args) {
6. Pattern p = Pattern.compile(args[0]);
7. Matcher m = p.matcher(args[1]);
8. int count = 0;
9. while(m.find())
10. count++;
11. System.out.print(count);
12. }
13. }
And given the command line invocation:
java Archie "\d+" ab2c4d67
What is the result?
A. 0
B. 3
C. 4
D. 8
E. 9
F. Compilation fails
Answer:
B is correct. The "\d" metacharacter looks for digits, and the + quantifier says look for "one or more" occurrences. The find() method will find three sets of one or more consecutive digits: 2, 4, and 67.
A, C, D, E, and F are incorrect based on the above. (Objective 3.5)

15. Given: 
import java.util.*;
public class Looking {
public static void main(String[] args) {
String input = "1 2 a 3 45 6";
Scanner sc = new Scanner(input);
int x = 0;
do {
x = sc.nextInt();
System.out.print(x + " ");
} while (x!=0);
}
}
What is the result?
A. 1 2
B. 1 2 3 45 6
C. 1 2 3 4 5 6
D. 1 2 a 3 45 6
E. Compilation fails
F. 1 2 followed by an exception
Answer:
F is correct. The nextXxx() methods are typically invoked after a call to a hasNextXxx(), which determines whether the next token is of the correct type.
A, B, C, D, and E are incorrect based on the above. (Objective 3.5)


.


regards,
TechDexters Support Team
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses
Posted by 0 comments

Aug 2, 2012

Operators


Relational Operators (Objective 7.6)
  1. Relational operators always result in a boolean value (true or false).
  2. There are six relational operators: >, >=, <, <=, ==, and !=. 
  3. The last two (== and !=) are sometimes referred to as equality operators.
  4. When comparing characters, Java uses the Unicode value of the character as the numerical value.
Equality operators
  1. There are two equality operators: == and != .
  2. Four types of things can be tested: numbers, characters, booleans, and reference variables.
  3. When comparing reference variables, == returns true only if both references refer to the same object. 
instanceof Operator (Objective 7.6)
  1. instanceof is for reference variables only, and checks for whether the object is of a particular type.
  2. The instanceof operator can be used only to test objects (or null) against class types that are in the same class hierarchy.
  3. For interfaces, an object passes the instanceof test if any of its superclasses implement the interface on the right side of the instanceof operator.
Arithmetic Operators (Objective 7.6)
  1. There are four primary math operators: add, subtract, multiply, and divide.
  2. The remainder operator (%), returns the remainder of a division.
  3. Expressions are evaluated from left to right, unless you add parentheses, or unless some operators in the expression have higher precedence than others.
  4. The *, /, and % operators have higher precedence than + and -.
String Concatenation Operator (Objective 7.6)
  1. If either operand is a String, the + operator concatenates the operands. 
  2. If both operands are numeric, the + operator adds the operands.
Increment/Decrement Operators (Objective 7.6)
  1. Prefix operators (++ and --) run before the value is used in the expression.
  2. Postfix operators (++ and --) run after the value is used in the expression.
  3. In any expression, both operands are fully evaluated before the operator is applied.
  4. Variables marked final cannot be incremented or decremented.
Ternary (Conditional Operator) (Objective 7.6)
  1. Returns one of two values based on whether a boolean expression is true or false.
  2. Returns the value after the ? if the expression is true.
  3. Returns the value after the : if the expression is false.
Logical Operators (Objective 7.6)
  1. The exam covers six "logical" operators: &, |, ^, !, &&, and ||.
  2. Logical operators work with two expressions (except for !) that must resolve to boolean values.
  3. The && and & operators return true only if both operands are true.
  4. The || and | operators return true if either or both operands are true.
  5. The && and || operators are known as short-circuit operators.
  6. The && operator does not evaluate the right operand if the left operand is false.
  7. The || does not evaluate the right operand if the left operand is true.
  8. The & and | operators always evaluate both operands.
  9. The ^ operator (called the "logical XOR"), returns true if exactly one operand is true.
  10. The ! operator (called the "inversion" operator), returns the opposite value of the boolean operand it precedes.


regards,
Tech Dexters Support Team,
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses

Posted by 0 comments

Jul 24, 2012

Basics


Bullet Points on some Basic Concepts:

Stack and Heap
  • Local variables (method variables) live on the stack.
  • Objects and their instance variables live on the heap.
Literals and Primitive Casting (Objective 1.3)
  • Integer literals can be decimal, octal (e.g. 013), or hexadecimal (e.g. 0x3d).
  • Literals for longs end in L or l.
  • Float literals end in F or f, double literals end in a digit or D or d.
  • The boolean literals are true and false.
  • Literals for chars are a single character inside single quotes: 'd'.
Scope (Objectives 1.3 and 7.6)
  • Scope refers to the lifetime of a variable.
  • There are four basic scopes:
    • Static variables live basically as long as their class lives.
    • Instance variables live as long as their object lives.
    • Local variables live as long as their method is on the stack; however, if their method invokes another method, they are temporarily unavailable.
    • Block variables (e.g., in a for or an if) live until the block completes.
Basic Assignments (Objectives 1.3 and 7.6)
  • Literal integers are implicitly ints.
  • Integer expressions always result in an int-sized result, never smaller.
  • Floating-point numbers are implicitly doubles (64 bits).
  • Narrowing a primitive truncates the high order bits.
  • Compound assignments (e.g. +=), perform an automatic cast.
  • A reference variable holds the bits that are used to refer to an object.
  • Reference variables can refer to subclasses of the declared type but not to superclasses.
  • When creating a new object, e.g., Button b = new Button(); three things happen:
    • Make a reference variable named b, of type Button
    • Create a new Button object
    • Assign the Button object to the reference variable b 
Using a Variable or Array Element That Is Uninitialized and Unassigned (Objectives 1.3 and 7.6)
  • When an array of objects is instantiated, objects within the array are not instantiated automatically, but all the references get the default value of null.
  • When an array of primitives is instantiated, elements get default values.
  • Instance variables are always initialized with a default value.
  • Local/automatic/method variables are never given a default value. If you attempt to use one before initializing it, you'll get a compiler error.
Passing Variables into Methods (Objective 7.3)
  • Methods can take primitives and/or object references as arguments.
  • Method arguments are always copies.
  • Method arguments are never actual objects (they can be references to objects).
  • A primitive argument is an unattached copy of the original primitive.
  • A reference argument is another copy of a reference to the original object.
  • Shadowing occurs when two variables with different scopes share the same name. This leads to hard-to-find bugs, and hard-to-answer exam questions.
Array Declaration, Construction, and Initialization (Obj. 1.3)
  • Arrays can hold primitives or objects, but the array itself is always an object.
  • When you declare an array, the brackets can be left or right of the name.
  • It is never legal to include the size of an array in the declaration.
  • You must include the size of an array when you construct it (using new) unless you are creating an anonymous array.
  • Elements in an array of objects are not automatically created, although primitive array elements are given default values.
  • You'll get a NullPointerException if you try to use an array element in an object array, if that element does not refer to a real object.
  • Arrays are indexed beginning with zero.
  • An ArrayIndexOutOfBoundsException occurs if you use a bad index value.
  • Arrays have a length variable whose value is the number of array elements.
  • The last index you can access is always one less than the length of the array.
  • Multidimensional arrays are just arrays of arrays.
  • The dimensions in a multidimensional array can have different lengths.
  • An array of primitives can accept any value that can be promoted implicitly to the array's declared type;. e.g., a byte variable can go in an int array.
  • An array of objects can hold any object that passes the IS-A (or instanceof) test for the declared type of the array. For example, if Horse extends Animal, then a Horse object can go into an Animal array.
  • If you assign an array to a previously declared array reference, the array you're assigning must be the same dimension as the reference you're assigning it to.
  • You can assign an array of one type to a previously declared array reference of one of its supertypes. For example, a Honda array can be assigned to an array declared as type Car (assuming Honda extends Car).
Initialization Blocks (Objectives 1.3 and 7.6)
  • Static initialization blocks run once, when the class is first loaded.
  • Instance initialization blocks run every time a new instance is created. They run after all super-constructors and before the constructor's code has run.
  • If multiple init blocks exist in a class, they follow the rules stated above, AND they run in the order in which they appear in the source file.
Using Wrappers (Objective 3.1)
  • The wrapper classes correlate to the primitive types.
  • Wrappers have two main functions:
    • To wrap primitives so that they can be handled like objects
    • To provide utility methods for primitives (usually conversions)
  • The three most important method families are
    • xxxValue() Takes no arguments, returns a primitive
    • parseXxx() Takes a String, returns a primitive, throws NFE
    • valueOf() Takes a String, returns a wrapped object, throws NFE
  • Wrapper constructors can take a String or a primitive, except for Character, which can only take a char.
  • Radix refers to bases (typically) other than 10; octal is radix = 8, hex = 16. 
Boxing (Objective 3.1)
  • As of Java 5, boxing allows you to convert primitives to wrappers or to convert wrappers to primitives automatically.
  • Using == with wrappers created through boxing is tricky; those with the same small values (typically lower than 127), will be ==, larger values will not be ==.
Advanced Overloading (Objectives 1.5 and 5.4)
  • Primitive widening uses the "smallest" method argument possible.
  • Used individually, boxing and var-args are compatible with overloading.
  • You CANNOT widen from one wrapper type to another. (IS-A fails.)
  • You CANNOT widen and then box. (An int can't become a Long.)
  • You can box and then widen. (An int can become an Object, via an Integer.)
  • You can combine var-args with either widening or boxing.
Garbage Collection (Objective 7.4)
  • In Java, garbage collection (GC) provides automated memory management.
  • The purpose of GC is to delete objects that can't be reached.
  • Only the JVM decides when to run the GC, you can only suggest it.
  • You can't know the GC algorithm for sure.
  • Objects must be considered eligible before they can be garbage collected.
  • An object is eligible when no live thread can reach it.
  • To reach an object, you must have a live, reachable reference to that object.
  • Java applications can run out of memory.
  • Islands of objects can be GCed, even though they refer to each other.
  • Request garbage collection with System.gc();
  • Class Object has a finalize() method.
  • The finalize() method is guaranteed to run once and only once before the garbage collector deletes an object.
  • The garbage collector makes no guarantees, finalize() may never run.
  • You can uneligibilize an object for GC from within finalize().

regards,
Tech Dexters Support Team,
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses

Posted by 0 comments

Jul 18, 2012

OCPJ Exercise-2 (Object Orientation)




1. Given: 
public abstract interface Frobnicate { public void twiddle(String s); }
Which is a correct class? (Choose all that apply.) A. 
public abstract class Frob implements Frobnicate {
public abstract void twiddle(String s) { }
}
B.
 public abstract class Frob implements Frobnicate { }
C.
public class Frob extends Frobnicate {
  public void twiddle(Integer i) { }
}
D. 
public class Frob implements Frobnicate {
public void twiddle(Integer i) { }
}
E. 
public class Frob implements Frobnicate {
  public void twiddle(String i) { }
  public void twiddle(Integer s) { }
}
Answer:
B is correct, an abstract class need not implement any or all of an interface's methods.
E is correct, the class implements the interface method and additionally overloads the twiddle() method.
A is incorrect because abstract methods have no body. C is incorrect because classes implement interfaces they don't extend them. D is incorrect because overloading a method is not implementing it.
(Objective 5.4)


2. Given: 
class Top {
   public Top(String s) { System.out.print("B"); }
}
public class Bottom2 extends Top {
   public Bottom2(String s) { System.out.print("D"); }
   public static void main(String [] args) {
     new Bottom2("C");
     System.out.println(" ");
   } 
}
What is the result?
A. BD
B. DB
C. BDC
D. DBC
E. Compilation fails
Answer:
E is correct. The implied super() call in Bottom2's constructor cannot be satisfied because there isn't a no-arg constructor in Top. A default, no-arg constructor is generated by the compiler only if the class has no constructor defined explicitly.
A, B, C, and D are incorrect based on the above. (Objective 1.6)


3. Given: 
class Clidder {
  private final void flipper() { System.out.println("Clidder"); }
}
public class Clidlet extends Clidder {
  public final void flipper() { System.out.println("Clidlet"); }
  public static void main(String [] args) {
  new Clidlet().flipper();
  }
}
What is the result? A. Clidlet
B. Clidder
C. Clidder Clidlet
D. Clidlet Clidder
E. Compilation fails
Answer:
A is correct. Although a final method cannot be overridden, in this case, the method is private, and therefore hidden. The effect is that a new, accessible, method flipper is created. Therefore, no polymorphism occurs in this example, the method invoked is simply that of the child class, and no error occurs.
B, C, D, and E are incorrect based on the preceding.(Objective 5.3)


4. Using the fragments below, complete the following code so it compiles. Note, you may not have to fill all of the slots.
Code: 
class AgedP {
   __________ __________ __________ __________ __________
   public AgedP(int x) {
       __________ __________ __________ __________ __________
   }
}

public class Kinder extends AgedP {
   __________ __________ __________ _________ ________ __________
   public Kinder(int x) {
     __________ __________ __________ __________ __________ ();
   }
}
Fragments: Use the following fragments zero or more times:
AgedP super this ( ) { } ;
Answer: 
class AgedP {
AgedP() {}
public AgedP(int x) {
}
}
public class Kinder extends AgedP {
public Kinder(int x) {
super();
}
}
As there is no droppable tile for the variable x and the parentheses (in the Kinder constructor), are already in place and empty, there is no way to construct a call to the superclass constructor that takes an argument. Therefore, the only remaining possibility is to create a call to the noargument superclass constructor. This is done as: super();. The line cannot be left blank, as the parentheses are already in place. Further, since the superclass constructor called is the noargument version, this constructor must be created. It will not be created by the compiler because there is another constructor already present.(Objective 5.4)


5 Which statement(s) are true? (Choose all that apply.) A. Cohesion is the OO principle most closely associated with hiding implementation details
B. Cohesion is the OO principle most closely associated with making sure that classes know about other classes only through their APIs
C. Cohesion is the OO principle most closely associated with making sure that a class is designed with a single, well-focused purpose
D. Cohesion is the OO principle most closely associated with allowing a single object to be seen as having many types
Answer:
Answer C is correct.
A refers to encapsulation, B refers to coupling, and D refers to polymorphism.(Objective 5.1)


6. Given the following, 
class X { void do1() { } }
class Y extends X { void do2() { } }

class Chrome {
  public static void main(String [] args) {
    X x1 = new X();
    X x2 = new Y();
    Y y1 = new Y();
    
    // insert code here
  }
}
Which, inserted at line 9, will compile? (Choose all that apply.)
A. x2.do2();
B. (Y)x2.do2();
C. ((Y)x2).do2();
D. None of the above statements will compile
Answer:
C is correct. Before you can invoke Y's do2 method you have to cast x2 to be of type Y. Statement B looks like a proper cast but without the second set of parentheses, the compiler thinks it's an incomplete statement.
A, B and D are incorrect based on the preceding.(Objective 5.2)


7. Given:
1. ClassA has a ClassD
2. Methods in ClassA use public methods in ClassB
3. Methods in ClassC use public methods in ClassA
4. Methods in ClassA use public variables in ClassB
Which is most likely true? (Choose the most likely.) A. ClassD has low cohesion
B. ClassA has weak encapsulation
C. ClassB has weak encapsulation
D. ClassB has strong encapsulation
E. ClassC is tightly coupled to ClassA
Answer:
C is correct. Generally speaking, public variables are a sign of weak encapsulation.
A, B, D, and E are incorrect, because based on the information given, none of these statements can be supported. (Objective 5.1)


8. Given: 
class Dog {
  public void bark() { System.out.print("woof "); }
}
class Hound extends Dog {
  public void sniff() { System.out.print("sniff "); }
  public void bark() { System.out.print("howl "); }
}
public class DogShow {
  public static void main(String[] args) { new DogShow().go(); }
  void go() {
    new Hound().bark();
    ((Dog) new Hound()).bark();
    ((Dog) new Hound()).sniff();
  }
}
What is the result? (Choose all that apply.)
A. howl howl sniff
B. howl woof sniff
C. howl howl followed by an exception
D. howl woof followed by an exception
E. Compilation fails with an error at line 12
F. Compilation fails with an error at line 13
Answer:
F is correct. Class Dog doesn't have a sniff method.
A, B, C, D, and E are incorrect based on the above information. (Objective 5.2)


9. Given: 
public class Redwood extends Tree {
   public static void main(String[] args) {
     new Redwood().go();
   }
   void go() {
     go2(new Tree(), new Redwood());
     go2((Redwood) new Tree(), new Redwood());
   }
   void go2(Tree t1, Redwood r1) {
     Redwood r2 = (Redwood)t1;
     Tree t2 = (Tree)r1;
   }
}
class Tree { }
What is the result? (Choose all that apply.)
A. An exception is thrown at runtime
B. The code compiles and runs with no output
C. Compilation fails with an error at line 6
D. Compilation fails with an error at line 7
E. Compilation fails with an error at line 10
F. Compilation fails with an error at line 11
Answer:
A is correct, a ClassCastException will be thrown when the code attempts to downcast a Tree to a Redwood.
B, C, D, E, and F are incorrect based on the above information.(Objective 5.2)


10. Given: 
public class Tenor extends Singer {
   public static String sing() { return "fa"; }
   public static void main(String[] args) {
     Tenor t = new Tenor();
     Singer s = new Tenor();
     System.out.println(t.sing() + " " + s.sing());
   }
}
class Singer { public static String sing() { return "la"; } }
What is the result?
A. fa fa
B. fa la
C. la la
D. Compilation fails
E. An exception is thrown at runtime
Answer:
B is correct. The code is correct, but polymorphism doesn't apply to static methods.
A, C, D, and E are incorrect based on the above information.(Objective 5.2)


11. Given: 
class Alpha {
  static String s = " ";
  protected Alpha() { s += "alpha "; }
}
class SubAlpha extends Alpha {
  private SubAlpha() { s += "sub "; }
}
public class SubSubAlpha extends Alpha {
  private SubSubAlpha() { s += "subsub "; }
  public static void main(String[] args) {
    new SubSubAlpha();
    System.out.println(s);
  }
}
What is the result? A. subsub
B. sub subsub
C. alpha subsub
D. alpha sub subsub
E. Compilation fails
F. An exception is thrown at runtime
Answer:
C is correct. Watch out, SubSubAlpha extends Alpha! Since the code doesn't attempt to make a SubAlpha, the private constructor in SubAlpha is okay.
A, B, D, E, and F are incorrect based on the above information. (Objective 5.3)


12. Given: 
class Building {
   Building() { System.out.print("b "); }
   Building(String name) {
      this(); 
      System.out.print("bn " + name);
   }
}
public class House extends Building {
   House() { System.out.print("h "); }
   House(String name) {
      this(); 
      System.out.print("hn " + name);
   }
   public static void main(String[] args) { new House("x "); }
}
what is the result?
A. h hn x
B. hn x h
C. b h hn x
D. b hn x h
E. bn x h hn x
F. b bn x h hn x
G. bn x b h hn x
H. Compilation fails
Answer:
C is correct. Remember that constructors call their superclass constructors, which execute first, and that constructors can be overloaded.
A, B, D, E, F, G, and H are incorrect based on the above information.(Objectives 1.6, 5.4)


13. Given: 
class Mammal {
String name = "furry ";
String makeNoise() { return "generic noise"; }
}
class Zebra extends Mammal {
String name = "stripes ";
String makeNoise() { return "bray"; }
}
public class ZooKeeper {
public static void main(String[] args) { new ZooKeeper().go(); }
void go() {
Mammal m = new Zebra();
System.out.println(m.name + m.makeNoise());
}
}
What is the result?
A. furry bray
B. stripes bray
C. furry generic noise
D. stripes generic noise
E. Compilation fails
F. An exception is thrown at runtime
Answer:
A is correct. Polymorphism is only for instance methods.
B, C, D, E, and F are incorrect based on the above information.(Objectives 1.5, 5.4)


14. You're designing a new online board game in which Floozels are a type of Jammers, Jammers can have Quizels, Quizels are a type of Klakker, and Floozels can have several Floozets. Which of the following fragments represent this design? (Choose all that apply.)
A. 
import java.util.*;
interface Klakker { }
class Jammer { Set q; }
class Quizel implements Klakker { }
public class Floozel extends Jammer { List f; }
interface Floozet { }
B. 
import java.util.*;
class Klakker { Set q; }
class Quizel extends Klakker { }
class Jammer { List f; }
class Floozet extends Floozel { }
public class Floozel { Set k; }
C. 
import java.util.*;
class Floozet { }
class Quizel implements Klakker { }
class Jammer { List q; }
interface Klakker { }
class Floozel extends Jammer { List f; }
D. 
import java.util.*;
interface Jammer extends Quizel { }
interface Klakker { }
interface Quizel extends Klakker { }
interface Floozel extends Jammer, Floozet { }
interface Floozet { }
Answer:
A and C are correct. The phrase "type of" indicates an "is-a" relationship (extends or implements), and the phrase “have” is of course a "has-a" relationship (usually instance variables).
B and D are incorrect based on the above information. (Objective 5.5)


15. Given: 
class A { }

class B extends A { }

public class ComingThru {
  static String s = "-";
  
  public static void main(String[] args) {
  A[] aa = new A[2];
  B[] ba = new B[2];
  sifter(aa);
  sifter(ba);
  sifter(7);
  
  System.out.println(s);
  }
  
  static void sifter(A[]... a2) { s += "1"; }
  static void sifter(B[]... b1) { s += "2"; }
  static void sifter(B[] b1) { s += "3"; }
  static void sifter(Object o) { s += "4"; }
}
What is the result?
A. -124
B. -134
C. -424
D. -434
E. -444
F. Compilation fails
Answer:
D is correct. In general, overloaded var-args methods are chosen last. Remember that arrays are objects. Finally, an int can be boxed to an Integer and then "widened" to an Object.
A, B, C, E, and F are incorrect based on the above information.(Objective 1.5)


regards,
Tech Dexters Support Team,
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses
Posted by 0 comments

Jul 11, 2012

Object Orientation


Encapsulation, IS-A, HAS-A (Objective 5.1)

  1. Encapsulation helps hide implementation behind an interface (or API).
  2. Encapsulated code has two features:
    • Instance variables are kept protected (usually with the private modifier).
    • Getter and setter methods provide access to instance variables.
  3. IS-A refers to inheritance or implementation.
  4. IS-A is expressed with the keyword extends.
  5. IS-A, "inherits from," and "is a subtype of " are all equivalent expressions.
  6. HAS-A means an instance of one class "has a" reference to an instance of another class or another instance of the same class.

Inheritance (Objective 5.5)

  1. Inheritance allows a class to be a subclass of a superclass, and thereby inherit public and protected variables and methods of the superclass.
  2. Inheritance is a key concept that underlies IS-A, polymorphism, overriding, overloading, and casting.
  3. All classes (except class Object), are subclasses of type Object, and therefore they inherit Object's methods.

Polymorphism (Objective 5.2)

  1. Polymorphism means "many forms."
  2. A reference variable is always of a single, unchangeable type, but it can refer to a subtype object.
  3. A single object can be referred to by reference variables of many different types —as long as they are the same type or a supertype of the object.
  4. The reference variable's type (not the object's type), determines which methods can be called!
  5. Polymorphic method invocations apply only to overridden instance methods.

Overriding and Overloading (Objectives 1.5 and 5.4)

  1. Methods can be overridden or overloaded; constructors can be overloaded but not overridden.
  2. Abstract methods must be overridden by the first concrete (non-abstract) subclass.
  3. With respect to the method it overrides, the overriding method
    • Must have the same argument list.
    • Must have the same return type, except that as of Java 5, the return type can be a subclass—this is known as a covariant return.
    • Must not have a more restrictive access modifier.
    • May have a less restrictive access modifier.
    • Must not throw new or broader checked exceptions.
    • May throw fewer or narrower checked exceptions, or any unchecked exception.
  4. final methods cannot be overridden.
  5. Only inherited methods may be overridden, and remember that private methods are not inherited.
  6. A subclass uses super.overriddenMethodName() to call the superclass version of an overridden method.
  7. Overloading means reusing a method name, but with different arguments.
  8. Overloaded methods
    • Must have different argument lists
    • May have different return types, if argument lists are also different
    • May have different access modifiers
    • May throw different exceptions
  9. Methods from a superclass can be overloaded in a subclass.
  10. Polymorphism applies to overriding, not to overloading.
  11. Object type (not the reference variable's type), determines which overridden method is used at runtime.
  12. Reference type determines which overloaded method will be used at compile time.

Reference Variable Casting (Objective 5.2)

  1. There are two types of reference variable casting: downcasting and upcasting.
  2. Downcasting: If you have a reference variable that refers to a subtype object, you can assign it to a reference variable of the subtype. You must make anexplicit cast to do this, and the result is that you can access the subtype's members with this new reference variable.
  3. Upcasting: You can assign a reference variable to a supertype reference variable explicitly or implicitly. This is an inherently safe operation because the assignment restricts the access capabilities of the new variable. 

Implementing an Interface (Objective 1.2)

  1. When you implement an interface, you are fulfilling its contract.
  2. You implement an interface by properly and concretely overriding all of the methods defined by the interface.
  3. A single class can implement many interfaces.

Return Types (Objective 1.5)

  1. Overloaded methods can change return types; overridden methods cannot, except in the case of covariant returns.
  2. Object reference return types can accept null as a return value.
  3. An array is a legal return type, both to declare and return as a value.
  4. For methods with primitive return types, any value that can be implicitly converted to the return type can be returned.
  5. Nothing can be returned from a void, but you can return nothing. You're allowed to simply say return, in any method with a void return type, to bust out of a method early. But you can't return nothing from a method with a non-void return type.
  6. Methods with an object reference return type, can return a subtype.
  7. Methods with an interface return type, can return any implementer.

Constructors and Instantiation (Objectives 1.6 and 5.4)

  1. A constructor is always invoked when a new object is created.
  2. Each superclass in an object's inheritance tree will have a constructor called.
  3. Every class, even an abstract class, has at least one constructor.
  4. Constructors must have the same name as the class.
  5. Constructors don't have a return type. If you see code with a return type, it's a method with the same name as the class, it's not a constructor.
  6. Typical constructor execution occurs as follows:
    • The constructor calls its superclass constructor, which calls its superclass constructor, and so on all the way up to the Object constructor.
    • The Object constructor executes and then returns to the calling constructor, which runs to completion and then returns to its calling constructor, and so on back down to the completion of the constructor of the actual instance being created.
  7. Constructors can use any access modifier (even private!).
  8. The compiler will create a default constructor if you don't create any constructors in your class.
  9. The default constructor is a no-arg constructor with a no-arg call to super().
  10. The first statement of every constructor must be a call to either this() (an overloaded constructor) or super().
  11. The compiler will add a call to super() unless you have already put in a call to this() or super().
  12. Instance members are accessible only after the super constructor runs.
  13. Abstract classes have constructors that are called when a concrete subclass is instantiated.
  14. Interfaces do not have constructors.
  15. If your superclass does not have a no-arg constructor, you must create a constructor and insert a call to super() with arguments matching those of the superclass constructor.
  16. Constructors are never inherited, thus they cannot be overridden. 
  17. A constructor can be directly invoked only by another constructor (using a call to super() or this()).
  18. Issues with calls to this()
    • May appear only as the first statement in a constructor.
    • The argument list determines which overloaded constructor is called. 
    • Constructors can call constructors can call constructors, and so on, but sooner or later one of them better call super() or the stack will explode.
    • Calls to this() and super() cannot be in the same constructor. You can have one or the other, but never both.

Statics (Objective 1.3)

  1. Use static methods to implement behaviors that are not affected by the state of any instances.
  2. Use static variables to hold data that is class specific as opposed to instance specific—there will be only one copy of a static variable.
  3. All static members belong to the class, not to any instance.
  4. A static method can't access an instance variable directly.
  5. static methods can't be overridden, but they can be redefined.
  6. Use the dot operator to access static members, but remember that using a reference variable with the dot operator is really a syntax trick, and the compiler will substitute the class name for the reference variable, for instance:

d.doStuff();
becomes:
Dog.doStuff();


Coupling and Cohesion (Objective 5.1)

  1. Coupling refers to the degree to which one class knows about or uses members of another class.
  2. Loose coupling is the desirable state of having classes that are well encapsulated, minimize references to each other, and limit the breadth of API usage.
  3. Tight coupling is the undesirable state of having classes that break the rules of loose coupling.
  4. Cohesion refers to the degree in which a class has a single, well-defined role or responsibility.
  5. High cohesion is the desirable state of a class whose members support a single, well-focused role or responsibility.
  6. Low cohesion is the undesirable state of a class whose members support multiple, unfocused roles or responsibilities.


regards,
Tech Dexters Support Team,
Hyderabad.


for Feedback and Support simply drop a mail in
techdexters{at}gmail.com
Remember to replace the {at} with @ in the email addresses

Posted by 2 comments
Subscribe to: Posts (Atom)