Ocjp Mock Jgi - II
- 1.
Given: 12. import java.io.*; 13. public class Forest implements Serializable { 14. private Tree tree = new Tree(); 15. public static void main(String [] args) { 16. Forest f = new Forest(); 17. try { 18. FileOutputStream fs = new FileOutputStream("Forest.ser"); 19. ObjectOutputStream os = new ObjectOutputStream(fs); 20. os.writeObject(f); os.close(); 21. } catch (Exception ex) { ex.printStackTrace(); } 22. } } 23. 24. class Tree { } What is the result?
- A.
Compilation fails.
- B.
An exception is thrown at runtime.
- C.
An instance of Forest is serialized.
- D.
An instance of Forest and an instance of Tree are both serialized.
Correct Answer
B. An exception is thrown at runtime.Explanation
The code is trying to serialize an instance of the class Forest. However, the class Tree does not implement the Serializable interface, so it cannot be serialized. This will result in a runtime exception being thrown.Rate this question:
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- 2.
Which capability exists only in java.io.FileWriter?
- A.
Closing an open stream.
- B.
Flushing an open stream.
- C.
Writing to an open stream.
- D.
Writing a line separator to an open stream.
Correct Answer
D. Writing a line separator to an open stream.Explanation
The capability that exists only in java.io.FileWriter is writing a line separator to an open stream. This means that FileWriter has a specific method or functionality that allows the user to write a line separator, which is a special character or sequence of characters used to indicate the end of a line, to an open stream. The other options mentioned, such as closing an open stream, flushing an open stream, and writing to an open stream, are capabilities that are generally available in various input/output classes in Java.Rate this question:
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- 3.
Given: 5. import java.io.*; 6. public class Talk { 7. public static void main(String[] args) { 8. Console c = new Console(); 9. String pw; 10. System.out.print("password: "); 11. pw = c.readLine(); 12. System.out.println("got " + pw); 13. } 14. } If the user types the password aiko when prompted, what is the result?
- A.
Password: got
- B.
Password: got aiko
- C.
Password: aiko got aiko
- D.
An exception is thrown at runtime.
- E.
Compilation fails due to an error on line 8.
Correct Answer
E. Compilation fails due to an error on line 8.Explanation
The code fails to compile due to an error on line 8 because the Console class is not imported properly. The correct import statement should be "import java.io.Console;" instead of "import java.io.*;". Therefore, the code cannot create an instance of the Console class and the compilation fails.Rate this question:
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- 4.
Given: 1. public class LineUp { 2. public static void main(String[] args) { 3. double d = 12.345; 4. // insert code here 5. } 6. } Which code fragment, inserted at line 4, produces the output | 12.345|?
- A.
System.out.printf("|%7d| \n", d);
- B.
System.out.printf("|%7f| \n", d);
- C.
System.out.printf("|%3.7d| \n", d);
- D.
System.out.printf("|%3.7f| \n", d);
- E.
System.out.printf("|%7.3d| \n", d);
- F.
System.out.printf("|%7.3f| \n", d);
Correct Answer
F. System.out.printf("|%7.3f| \n", d);Explanation
The code fragment "System.out.printf("|%7.3f| ", d);" produces the output "| 12.345|" because the "%7.3f" format specifier is used to format the double value "d" with a width of 7 characters, including the decimal point and any leading spaces. The ".3" specifies that the output should have 3 decimal places. The output is enclosed in "|" to match the desired format.Rate this question:
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- 5.
Given: 11. String test = "Test A. Test B. Test C."; 12. // insert code here 13. String[] result = test.split(regex); Which regular expression, inserted at line 12, correctly splits test into "Test A", "Test B", and "Test C"?
- A.
String regex = "";
- B.
String regex = " ";
- C.
String regex = ".*";
- D.
String regex = "\\s";
- E.
String regex = "\\.\\s*";
- F.
String regex = "\\w[ \.] +";
Correct Answer
E. String regex = "\\.\\s*";Explanation
The regular expression "\\.\\s*" correctly splits the string "test" into "Test A", "Test B", and "Test C". This regular expression matches a period followed by zero or more whitespace characters, which is the pattern used to separate the different parts of the string.Rate this question:
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- 6.
Given: 1. interface A { public void aMethod(); } 2. interface B { public void bMethod(); } 3. interface C extends A,B { public void cMethod(); } 4. class D implements B { 5. public void bMethod(){} 6. } 7. class E extends D implements C { 8. public void aMethod(){} 9. public void bMethod(){} 10. public void cMethod(){} 11. } What is the result?
- A.
Compilation fails because of an error in line 3.
- B.
Compilation fails because of an error in line 7.
- C.
Compilation fails because of an error in line 9.
- D.
If you define D e = new E(), then e.bMethod() invokes the version of bMethod() defined in Line 5.
- E.
If you define D e = (D)(new E()), then e.bMethod() invokes the version of bMethod() defined in Line 5.
- F.
If you define D e = (D)(new E()), then e.bMethod() invokes the version of bMethod() defined in Line 9.
Correct Answer
F. If you define D e = (D)(new E()), then e.bMethod() invokes the version of bMethod() defined in Line 9.Explanation
The correct answer is "If you define D e = (D)(new E()), then e.bMethod() invokes the version of bMethod() defined in Line 9." This is because class E extends class D, which implements interface B. Therefore, class E inherits the implementation of bMethod() from class D. In line 9, class E overrides the implementation of bMethod() from class D, so when invoking e.bMethod(), it will invoke the version of bMethod() defined in line 9.Rate this question:
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- 7.
What is the result ? 1. public class SimplCalc { 2. public int value; 3. public void calculate() { value += 7; } 4. } And: 1. public class MultiCalc extends SimpleCalc { 2. public void calculate() { value -=3; } 3. public void calculate(int multiplier) { 4. calculate(); 5. super.calculate(); 6. value *= multiplier; 7. } 8. public static void main(String[] args) { 9. MultiCalc calculator = new MultiCalc(); 10. calculator.calculate(2); 11. System.out.println("Value is: "+calculator.value); 12. } 13. }
- A.
Value is: 8
- B.
Compilation fails.
- C.
Value is: 12
- D.
Value is: -12
- E.
The code runs with no output.
- F.
An exception is thrown at runtime.
Correct Answer
A. Value is: 8Explanation
The correct answer is "Value is: 8" because the code creates an instance of the MultiCalc class and calls the calculate() method with an argument of 2. Inside the calculate() method, the calculate() method is called again without any arguments, which adds 7 to the value. Then, the super.calculate() method is called, which adds another 7 to the value. Finally, the value is multiplied by the multiplier, which is 2. Therefore, the final value is 8.Rate this question:
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- 8.
Given: 1. public class Base { 2. public static final String FOO = "foo"; 3. public static void main(String[] args) { 4. Base b = new Base(); 5. Sub s = new Sub(); 6. System.out.print(Base.FOO); 7. System.out.print(Sub.FOO); 8. System.out.print(b.FOO); 9. System.out.print(s.FOO); 10. System.out.print(((Base)s).FOO); 11. } } 12. class Sub extends Base {public static final String FOO="bar";} What is the result?
- A.
Foofoofoofoofoo
- B.
Foobarfoobarbar
- C.
Foobarfoofoofoo
- D.
Foobarfoobarfoo
- E.
Barbarbarbarbar
- F.
Foofoofoobarbar
- G.
Foofoofoobarfoo
Correct Answer
D. FoobarfoobarfooExplanation
The correct answer is "foobarfoobarfoo".
In this code, the class "Base" has a public static final String variable "FOO" with the value "foo". The class "Sub" extends "Base" and also has a public static final String variable "FOO" with the value "bar".
In the main method, the code creates an instance of "Base" and an instance of "Sub".
When printing the value of "FOO" for each object, the output is as follows:
- Base.FOO: "foo"
- Sub.FOO: "bar"
- b.FOO: "foo" (since it is referencing the "FOO" variable of the "Base" class)
- s.FOO: "bar" (since it is referencing the "FOO" variable of the "Sub" class)
- ((Base)s).FOO: "foo" (since it is casting "s" to the "Base" class and referencing the "FOO" variable of the "Base" class)
Therefore, the final output is "foobarfoobarfoo".Rate this question:
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- 9.
Given: 11. class Mammal { } 12. 13. class Raccoon extends Mammal { 14. Mammal m = new Mammal(); 15. } 16. 17. class BabyRaccoon extends Mammal { } Which four statements are true? (Choose four.)
- A.
Raccoon is-a Mammal.
- B.
Raccoon has-a Mammal.
- C.
BabyRaccoon is-a Mammal.
- D.
BabyRaccoon is-a Raccoon.
- E.
BabyRaccoon has-a Mammal.
- F.
BabyRaccoon is-a BabyRaccoon.
Correct Answer(s)
A. Raccoon is-a Mammal.
B. Raccoon has-a Mammal.
C. BabyRaccoon is-a Mammal.
F. BabyRaccoon is-a BabyRaccoon.Explanation
The first statement, "Raccoon is-a Mammal," is true because the class Raccoon extends the class Mammal, indicating that Raccoon is a type of Mammal. The second statement, "Raccoon has-a Mammal," is false because the Raccoon class does not have a Mammal object as a member variable. The third statement, "BabyRaccoon is-a Mammal," is true because the class BabyRaccoon extends the class Mammal, indicating that BabyRaccoon is a type of Mammal. The fourth statement, "BabyRaccoon is-a BabyRaccoon," is true because a class is always considered to be an instance of itself.Rate this question:
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- 10.
Given: 10. interface A { void x(); } 11. class B implements A { public void x() {} public void y() {} } 12. class C extends B { public void x() {} } And: 20. java.util.List<A> list = new java.util.ArrayList<A>(); 21. list.add(new B()); 22. list.add(new C()); 23. for (A a : list) { 24. a.x(); 25. a.y(); 26. } What is the result?
- A.
The code runs with no output.
- B.
An exception is thrown at runtime.
- C.
Compilation fails because of an error in line 20.
- D.
Compilation fails because of an error in line 21.
- E.
Compilation fails because of an error in line 23.
- F.
Compilation fails because of an error in line 25.
Correct Answer
F. Compilation fails because of an error in line 25.Explanation
The code fails to compile because the method "y()" is not defined in the interface A. Since the variable "a" is of type A, it does not have access to the method "y()" defined in class B. Therefore, calling "a.y()" on line 25 would result in a compilation error.Rate this question:
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- 11.
Given: 2. public class Hi { 3. void m1() { } 4. protected void() m2 { } 5. } 6. class Lois extends Hi { 7. // insert code here 8. } Which four code fragments, inserted independently at line 7, will compile? (Choose four.)
- A.
Public void m1() { }
- B.
Protected void m1() { }
- C.
Private void m1() { }
- D.
Void m2() { }
- E.
Public void m2() { }
- F.
Protected void m2() { }
- G.
Private void m2() { }
Correct Answer(s)
A. Public void m1() { }
B. Protected void m1() { }
E. Public void m2() { }
F. Protected void m2() { }Explanation
The given code is a class named "Hi" with two methods, "m1()" and "m2()", both without any return type. The class "Lois" extends the "Hi" class.
To compile the code at line 7, we need to insert code that is valid and does not cause any compilation errors.
The four code fragments that can be inserted at line 7 and will compile are:
- public void m1() { }
- protected void m1() { }
- public void m2() { }
- protected void m2() { }Rate this question:
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- 12.
Which four statements are true? (Choose four.)
- A.
Has-a relationships should never be encapsulated.
- B.
Has-a relationships should be implemented using inheritance.
- C.
Has-a relationships can be implemented using instance variables.
- D.
Is-a relationships can be implemented using the extends keyword.
- E.
Is-a relationships can be implemented using the implements keyword.
- F.
The relationship between Movie and Actress is an example of an is-a relationship.
- G.
An array or a collection can be used to implement a one-to-many has-a relationship.
Correct Answer(s)
C. Has-a relationships can be implemented using instance variables.
D. Is-a relationships can be implemented using the extends keyword.
E. Is-a relationships can be implemented using the implements keyword.
G. An array or a collection can be used to implement a one-to-many has-a relationship.Explanation
Has-a relationships can be implemented using instance variables because an object can have another object as one of its instance variables. Is-a relationships can be implemented using the extends keyword because a class can inherit from another class to establish an is-a relationship. Is-a relationships can also be implemented using the implements keyword to establish a relationship between a class and an interface. An array or a collection can be used to implement a one-to-many has-a relationship because it allows multiple objects to be stored and accessed as a single entity.Rate this question:
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- 13.
Given: 10: public class Hello { 11: String title; 12: int value; 13: public Hello() { 14: title += " World"; 15: } 16: public Hello(int value) { 17: this.value = value; 18: title = "Hello"; 19: Hello(); 20: } 21: } and: 30: Hello c = new Hello(5); 31: System.out.println(c.title); What is the result?
- A.
Hello
- B.
Hello World
- C.
Compilation fails
- D.
Hello World 5
- E.
The code runs with no output
- F.
An exception is thrown at runtime
Correct Answer
C. Compilation failsExplanation
The code fails to compile because the constructor on line 19 is calling the constructor on line 13 using the syntax "Hello();". However, this syntax is not valid for calling a constructor within another constructor. The correct syntax would be "this();" to call the default constructor.Rate this question:
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- 14.
Given: 1. package geometry; 2. public class Hypotenuse { 3. public InnerTriangle it = new InnerTriangle(); 4. class InnerTriangle { 5. public int base; 6. public int height; 7. } 8. } Which statement is true about the class of an object that can reference the variable base?
- A.
It can be any class.
- B.
No class has access to base.
- C.
The class must belong to the geometry package.
- D.
The class must be a subclass of the class Hypotenuse.
Correct Answer
C. The class must belong to the geometry package.Explanation
The variable "base" is a public variable in the InnerTriangle class, which is a member of the geometry package. Therefore, any class that wants to access the variable "base" must also belong to the geometry package.Rate this question:
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- 15.
Given this code from Class B: 25. A a1 = new A(); 26. A a2 = new A(); 27. A a3 = new A(); 28. System.out.println(A.getInstanceCount()); What is the result? 1. public class A { 2. 3. private int counter = 0; 4. 5. public static int getInstanceCount() { 6. return counter; 7. } 8. 9. public A() { 10. counter++; 11. } 12. 13. }
- A.
Compilation of class A fails.
- B.
Line 28 prints the value 3 to System.out.
- C.
Line 28 prints the value 1 to System.out.
- D.
A runtime error occurs when line 25 executes.
- E.
Compilation fails because of an error on line 28.
Correct Answer
A. Compilation of class A fails.Explanation
The code in class A fails to compile because the variable "counter" is declared as a non-static variable, but it is accessed in a static method "getInstanceCount()". To fix this error, the variable "counter" should be declared as static.Rate this question:
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- 16.
Given: 10. interface Data { public void load(); } 11. abstract class Info { public abstract void load(); } Which class correctly uses the Data interface and Info class?
- A.
Public class Employee extends Info implements Data { public void load() { /*do something*/ } }
- B.
Public class Employee implements Info extends Data { public void load() { /*do something*/ } }
- C.
Public class Employee extends Info implements Data public void load(){ /*do something*/ } public void Info.load(){ /*do something*/ } }
- D.
Public class Employee implements Info extends Data { public void Data.load(){ /*do something*/ } public void load(){ /*do something*/ } }
- E.
Public class Employee implements Info extends Data { public void load(){ /*do something*/ } public void Info.load(){ /*do something*/ } }
- F.
Public class Employee extends Info implements Data{ public void Data.load() { /*do something*/ } public void Info.load() { /*do something*/ } }
Correct Answer
A. Public class Employee extends Info implements Data { public void load() { /*do something*/ } }Explanation
The correct answer is "public class Employee extends Info implements Data { public void load() { /*do something*/ } }". This is the correct class because it correctly implements the Data interface and extends the Info class. The load() method in this class satisfies the requirements of both the Data interface and the Info class.Rate this question:
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- 17.
Given: 1. class Alligator { 2. public static void main(String[] args) { 3. int []x[] = {{1,2}, {3,4,5}, {6,7,8,9}}; 4. int [][]y = x; 5. System.out.println(y[2][1]); 6. } 7. } What is the result?
- A.
2
- B.
3
- C.
4
- D.
6
- E.
7
- F.
Compilation fails
Correct Answer
E. 7Explanation
The code declares and initializes a two-dimensional array `x` with three rows and varying number of columns. Then, it assigns `x` to another two-dimensional array `y`. Finally, it prints the value at index `[2][1]` of `y`, which is `7`.Rate this question:
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- 18.
Given: 21. abstract class C1 { 22. public C1() { System.out.print(1); } 23. } 24. class C2 extends C1 { 25. public C2() { System.out.print(2); } 26. } 27. class C3 extends C2 { 28. public C3() { System.out.println(3); } 29. } 30. public class Ctest { 31. public static void main(String[] a) { new C3(); } 32. } What is the result?
- A.
3
- B.
23
- C.
32
- D.
123
- E.
321
- F.
Compilation fails.
- G.
An exception is thrown at runtime.
Correct Answer
D. 123Explanation
The code defines an abstract class C1 with a constructor that prints 1. Class C2 extends C1 and has a constructor that prints 2. Class C3 extends C2 and has a constructor that prints 3. In the main method of the Ctest class, a new object of C3 is created. When this object is created, the constructors of C1, C2, and C3 are called in order, resulting in the output 123.Rate this question:
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- 19.
Given: 10. class One { 11. public One foo() { return this; } 12. } 13. class Two extends One { 14. public One foo() { return this; } 15. } 16. class Three extends Two { 17. // insert method here 18. } Which two methods, inserted individually, correctly complete the Three class? (Choose two.)
- A.
Public void foo() {}
- B.
Public int foo() { return 3; }
- C.
Public Two foo() { return this; }
- D.
Public One foo() { return this; }
- E.
Public Object foo() { return this; }
Correct Answer(s)
C. Public Two foo() { return this; }
D. Public One foo() { return this; }Explanation
The correct answer is "public Two foo() { return this; }" and "public One foo() { return this; }". These two methods correctly complete the Three class because they override the foo() method from the parent class, Two, and return the appropriate type of object. The first method returns an object of type Two, which is the class that directly extends One. The second method returns an object of type One, which is the class that the Three class ultimately extends.Rate this question:
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- 20.
Which two classes correctly implement both the java.lang.Runnable and the java.lang. Cloneable interfaces? (Choose two.)
- A.
Public class Session implements Runnable, Cloneable { public void run(); public Object clone(); }
- B.
Public class Session extends Runnable, Cloneable { public void run() { /* do something */ } public Object clone() { /* make a copy */ }
- C.
Public class Session implements Runnable, Cloneable { public void run() { /* do something */ } public Object clone() { /* make a copy */ }
- D.
Public abstract class Session implements Runnable, Cloneable { public void run() { /* do something */ } public Object clone() { /*make a copy */ }
- E.
Public class Session implements Runnable, implements Cloneable { public void run() { /* do something */ } public Object clone() { /* make a copy */ }
Correct Answer(s)
C. Public class Session implements Runnable, Cloneable { public void run() { /* do something */ } public Object clone() { /* make a copy */ }
D. Public abstract class Session implements Runnable, Cloneable { public void run() { /* do something */ } public Object clone() { /*make a copy */ }Explanation
The correct answers are the first and fourth options. The first option correctly implements both the Runnable and Cloneable interfaces by providing the run() and clone() methods. The fourth option also correctly implements both interfaces, as it extends the abstract class Session and provides the necessary methods.Rate this question:
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- 21.
Given: 11. public interface A { public void m1(); } 12. 13. class B implements A { } 14. class C implements A { public void m1() { } } 15. class D implements A { public void m1(int x) { } } 16. abstract class E implements A { } 17. abstract class F implements A { public void m1() { } } 18. abstract class G implements A { public void m1(int x) { } } What is the result?
- A.
Compilation succeeds.
- B.
Exactly one class does NOT compile.
- C.
Exactly two classes do NOT compile.
- D.
Exactly four classes do NOT compile.
- E.
Exactly three classes do NOT compile.
Correct Answer
C. Exactly two classes do NOT compile.Explanation
The two classes that do not compile are class D and class G. Class D does not compile because it implements the interface A but does not provide an implementation for the method m1() without any parameters, as required by the interface. Class G does not compile for the same reason, as it also does not provide an implementation for the method m1() without any parameters. The other classes either implement the method m1() as required by the interface or are abstract classes that are not required to provide an implementation.Rate this question:
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- 22.
Given: 10. class Line { 11. public class Point { public int x,y;} 12. public Point getPoint() { return new Point(); } 13. } 14. class Triangle { 15. public Triangle() { 16. // insert code here 17. } 18. } Which code, inserted at line 16, correctly retrieves a local instance of a Point object?
- A.
Point p = Line.getPoint();
- B.
Line.Point p = Line.getPoint();
- C.
Point p = (new Line()).getPoint();
- D.
Line.Point p = (new Line()).getPoint();
Correct Answer
D. Line.Point p = (new Line()).getPoint();Explanation
The correct answer is "Line.Point p = (new Line()).getPoint();". This code correctly retrieves a local instance of a Point object by creating a new instance of the Line class using the "new Line()" syntax and then calling the "getPoint()" method on that instance. The "Line.Point" syntax is used to specify that the Point class is a nested class within the Line class.Rate this question:
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- 23.
Given: 1. class TestA { 2. public void start() { System.out.println("TestA"); } 3. } 4. public class TestB extends TestA { 5. public void start() { System.out.println("TestB"); } 6. public static void main(String[] args) { 7. ((TestA)new TestB()).start(); 8. } 9. } What is the result?
- A.
TestA
- B.
TestB
- C.
Compilation fails.
- D.
An exception is thrown at runtime.
Correct Answer
B. TestBExplanation
The code creates a class TestA with a method start() that prints "TestA". Then, a class TestB is created that extends TestA and overrides the start() method to print "TestB". In the main method, an instance of TestB is created and casted to TestA. The start() method is then called on this instance, which results in "TestB" being printed. Therefore, the result is "TestB".Rate this question:
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- 24.
Which statement is true about the classes and interfaces in the exhibit? 1. public interface A { 2. public void doSomething(String thing); 3. } 1. public class AImpl implements A { 2. public void doSomething(String thing){ } 3. } 1. public class B { 2. public A doIt() { 3. // more code here 4. } 5. 6. public String execute() { 7. // more code here 8. } 9. } 1. public class C extends B { 2. public AImpl doIt() { 3. // more code here 4. } 5. 6. public Object execute() { 7. // more code here 8. } 9. }
- A.
Compilation will succeed for all classes and interfaces.
- B.
Compilation of class C will fail because of an error in line 2.
- C.
Compilation of class C will fail because of an error in line 6.
- D.
Compilation of class AImpl will fail because of an error in line 2.
Correct Answer
C. Compilation of class C will fail because of an error in line 6.Explanation
The compilation of class C will fail because the method "execute()" in class C has a different return type (Object) than the method it is overriding in class B (String). The return type of an overridden method must be the same or a subtype of the return type in the superclass.Rate this question:
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- 25.
Given the following six method names: addListener addMouseListener setMouseListener deleteMouseListener removeMouseListener registerMouseListener How many of these method names follow JavaBean Listener naming rules?
- A.
1
- B.
2
- C.
3
- D.
4
- E.
5
Correct Answer
B. 2Explanation
Two of the given method names follow JavaBean Listener naming rules. The naming convention for JavaBean Listener methods is to start with "add" followed by the event type, and end with "Listener". In this case, the method names "addListener" and "addMouseListener" follow this convention. The other method names do not follow the convention as they either have incorrect prefixes or do not end with "Listener".Rate this question:
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- 26.
Given: 11. public static void main(String[] args) { 12. Object obj = new int[] { 1, 2, 3 }; 13. int[] someArray = (int[])obj; 14. for (int i : someArray) System.out.print(i + " "); 15. } What is the result?
- A.
1 2 3
- B.
Compilation fails because of an error in line 12.
- C.
Compilation fails because of an error in line 13.
- D.
Compilation fails because of an error in line 14.
- E.
A ClassCastException is thrown at runtime.
Correct Answer
A. 1 2 3Explanation
The correct answer is "1 2 3" because the code creates an array of integers on line 12 and assigns it to an Object reference. On line 13, the code casts the Object reference back to an int array. Then, on line 14, a for-each loop is used to iterate over the elements of the int array and print them out. Since the int array contains the values 1, 2, and 3, the output will be "1 2 3".Rate this question:
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- 27.
Given: foo and bar are public references available to many other threads. foo refers to a Thread and bar is an Object. The thread foo is currently executing bar.wait(). From another thread, what provides the most reliable way to ensure that foo will stop executing wait()?
- A.
Foo.notify();
- B.
Bar.notify();
- C.
Foo.notifyAll();
- D.
Thread.notify();
- E.
Bar.notifyAll();
- F.
Object.notify();
Correct Answer
E. Bar.notifyAll();Explanation
The most reliable way to ensure that foo will stop executing wait() is by calling bar.notifyAll(). This is because bar is the object on which the wait() method is being called, so calling notifyAll() on bar will notify all threads waiting on that object, including foo. This will allow foo to resume execution and continue its task.Rate this question:
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- 28.
1. public class Threads1 { 2. int x = 0; 3. public class Runner implements Runnable { 4. public void run() { 5. int current = 0; 6. for(int i=0;i<4;i++) { 7. current = x; 8. System.out.print(current + ", "); 9. x = current + 2; 10. } 11. } 12. } 13. 14. public static void main(String[] args) { 15. new Threads1().go(); 16. } 17. 18. public void go() { 19. Runnable r1 = new Runnable(); 20. new Thread(r1).start(); 21. new Thread(r1).start(); 22. } 23. } Which two are possible results? (Choose two.)
- A.
0, 2, 4, 4, 6, 8, 10, 6,
- B.
0, 2, 4, 6, 8, 10, 2, 4,
- C.
0, 2, 4, 6, 8, 10, 12, 14,
- D.
0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10, 10, 12, 12, 14, 14,
- E.
0, 2, 4, 6, 8, 10, 12, 14, 0, 2, 4, 6, 8, 10, 12, 14,
Correct Answer(s)
A. 0, 2, 4, 4, 6, 8, 10, 6,
C. 0, 2, 4, 6, 8, 10, 12, 14,Explanation
The code creates two threads that both execute the same Runnable object. The Runnable object has a for loop that runs four times. In each iteration, the current value of x is printed followed by a comma. Then, the value of x is updated by adding 2 to the current value. Since both threads are executing the same Runnable object, they will share the same value of x. Therefore, the possible results will be a combination of the printed values of x from both threads. In this case, the two possible results are 0, 2, 4, 4, 6, 8, 10, 6 and 0, 2, 4, 6, 8, 10, 12, 14.Rate this question:
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- 29.
What is the output if the main() method is run? 10. public class Starter extends Thread { 11. private int x = 2; 12. public static void main(String[] args)throws Exception { 13. new Starter().makeItSo(); 14. } 15. public Starter() { 16. x = 5; 17. start(); 18. } 19. public void makeItSo() throws Exception { 20. join(); 21. x = x - 1; 22. System.out.println(x); 23. } 24. public void run() { x *= 2; } 25. }
- A.
4
- B.
5
- C.
8
- D.
9
- E.
Compilation fails.
- F.
An exception is thrown at runtime.
- G.
It is impossible to determine for certain.
Correct Answer
D. 9Explanation
The output of the program is 9.
- The main method creates an instance of the Starter class and calls the makeItSo() method.
- In the constructor of the Starter class, the value of x is set to 5 and the start() method is called, which will eventually execute the run() method.
- The run() method multiplies the value of x by 2, resulting in x being 10.
- The makeItSo() method then calls the join() method, which waits for the thread to complete before continuing.
- After the join() method, the value of x is decremented by 1, resulting in x being 9.
- Finally, the value of x (which is 9) is printed to the console.Rate this question:
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- 30.
Given: 11. public class PingPong implements Runnable { 12. synchronized void hit(long n) { 13. for(int i = 1; i < 3; i++) 14. System.out.print(n + "-" + i + " "); 15. } 16. public static void main(String[] args) { 17. new Thread(new PingPong()).start(); 18. new Thread(new PingPong()).start(); 19. } 20. public void run() { 21. hit(Thread.currentThread().getId()); 22. } 23. } Which two statements are true? (Choose two.)
- A.
The output could be 8-1 7-2 8-2 7-1
- B.
The output could be 7-1 7-2 8-1 6-1
- C.
The output could be 8-1 7-1 7-2 8-2
- D.
The output could be 8-1 8-2 7-1 7-2
Correct Answer(s)
C. The output could be 8-1 7-1 7-2 8-2
D. The output could be 8-1 8-2 7-1 7-2Explanation
The output could be 8-1 7-1 7-2 8-2 because the hit() method is synchronized, which means only one thread can access it at a time. The two threads created in the main method start simultaneously and call the hit() method. However, only one thread can enter the hit() method at a time, so one thread will complete its execution before the other can enter. This leads to the possibility of the output being in any order depending on which thread enters the hit() method first.Rate this question:
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- 31.
1. class Computation extends Thread { 2. 3. private int num; 4. private boolean isComplete; 5. private int result; 6. 7. public Computation(int num) { this.num = num; } 8. 9. public synchronized void run() { 10. result = num * 2; 11. isComplete = true; 12. notify(); 13. } 14. 15. public synchronized ing getResult() { 16. while(!isComplete) { 17. try { 18. wait(); 19. } catch(InterruptedException e){ } 20. } 21. return result; 22. } 23. 24. public static void main(String[] args) { 25. Computation[] computations = new Computation[4]; 26. for(int i = 0; i < computations.length; i++) { 27. computations[i] = new Computation(i); 28. computations[i].start(); 29. } 30. for(Computation c:computations) 31. System.out.print(c.getResult()+" "); 32. } 33. } What is the result?
- A.
The code will deadlock.
- B.
The code may run with no output.
- C.
An exception is thrown at runtime.
- D.
The code may run with output "0 6".
- E.
The code may run with output "2 0 6 4".
- F.
The code may run with output "0 2 4 6".
Correct Answer
F. The code may run with output "0 2 4 6".Explanation
The code creates an array of Computation objects and starts each thread. Each thread calculates the result by multiplying the given number by 2 and sets the isComplete flag to true. The main thread then waits for each computation to complete and prints the result. Since the computations are running concurrently, the order in which they complete is not guaranteed. Therefore, the output can be "0 2 4 6" or any other possible permutation of these numbers.Rate this question:
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- 32.
Which two code fragments will execute the method doStuff() in a separate thread? (Choose two.)
- A.
New Thread() { public void run() { doStuff(); } };
- B.
New Thread() { public void start() { doStuff(); } };
- C.
New Thread() { public void start() { doStuff(); } }.run();
- D.
New Thread() { public void run() { doStuff(); } }.start();
- E.
New Thread(new Runnable() { public void run() { doStuff(); } }).run();
- F.
New Thread(new Runnable() { public void run() { doStuff(); } }).start();
Correct Answer(s)
D. New Thread() { public void run() { doStuff(); } }.start();
F. New Thread(new Runnable() { public void run() { doStuff(); } }).start();Explanation
The two code fragments that will execute the method doStuff() in a separate thread are:
1. new Thread() {
public void run() { doStuff(); }
}.start();
This code creates a new thread using the Thread class and overrides the run() method to call the doStuff() method. The start() method is then called to start the thread and execute the doStuff() method in a separate thread.
2. new Thread(new Runnable() {
public void run() { doStuff(); }
}).start();
This code creates a new thread using the Thread class and passes a Runnable object as a parameter. The run() method of the Runnable object is overridden to call the doStuff() method. The start() method is then called to start the thread and execute the doStuff() method in a separate thread.Rate this question:
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- 33.
Given: 11. public class Person { 12. private String name; 13. public Person(String name) { 14. this.name = name; 15. } 16. public boolean equals(Object o) { 17. if ( ! ( o instanceof Person) ) return false; 18. Person p = (Person) o; 19. return p.name.equals(this.name); 20. } 21. } Which statement is true?
- A.
Compilation fails because the hashCode method is not overridden.
- B.
A HashSet could contain multiple Person objects with the same name.
- C.
All Person objects will have the same hash code because the hashCode method is not overridden.
- D.
If a HashSet contains more than one Person object with name="Fred", then removing another Person, also with name="Fred", will remove them all.
Correct Answer
B. A HashSet could contain multiple Person objects with the same name.Explanation
The correct answer is that a HashSet could contain multiple Person objects with the same name. This is because the equals() method in the Person class is overridden to compare the names of two Person objects. However, the hashCode() method is not overridden, which means that the default implementation from the Object class is used. This default implementation generates a unique hash code for each object, regardless of the values of their fields. Therefore, multiple Person objects with the same name will have different hash codes and can be stored in a HashSet.Rate this question:
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- 34.
Given: 5. import java.util.*; 6. public class SortOf { 7. public static void main(String[] args) { 8. ArrayList<Integer> a = new ArrayList<Integer>(); 9. a.add(1); a.add(5); a.add(3); 11. Collections.sort(a); 12. a.add(2); 13. Collections.reverse(a); 14. System.out.println(a); 15. } 16. } What is the result?
- A.
[1, 2, 3, 5]
- B.
[2, 1, 3, 5]
- C.
[2, 5, 3, 1]
- D.
[5, 3, 2, 1]
- E.
[1, 3, 5, 2]
- F.
Compilation fails.
- G.
An exception is thrown at runtime.
Correct Answer
C. [2, 5, 3, 1]Explanation
The given code first creates an ArrayList named 'a' and adds integers 1, 5, and 3 to it. Then, it sorts the ArrayList using the Collections.sort() method, resulting in [1, 3, 5]. After that, it adds the integer 2 to the ArrayList using the add() method, resulting in [1, 3, 5, 2]. Finally, it reverses the order of the elements in the ArrayList using the Collections.reverse() method, resulting in [2, 5, 3, 1]. Therefore, the correct answer is [2, 5, 3, 1].Rate this question:
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- 35.
Given: 11. public class Person { 12. private name; 13. public Person(String name) { 14. this.name = name; 15. } 16. public int hashCode() { 17. return 420; 18. } 19. } Which statement is true?
- A.
The time to find the value from HashMap with a Person key depends on the size of the map.
- B.
Deleting a Person key from a HashMap will delete all map entries for all keys of type Person.
- C.
Inserting a second Person object into a HashSet will cause the first Person object to be removed as a duplicate.
- D.
The time to determine whether a Person object is contained in a HashSet is constant and does NOT depend on the size of the map.
Correct Answer
A. The time to find the value from HashMap with a Person key depends on the size of the map.Explanation
The given answer is correct because in a HashMap, the time to find the value associated with a specific key depends on the size of the map. As the size of the map increases, the time to find the value also increases because the HashMap uses the hash code of the key to determine the bucket where the value is stored. If the number of elements in the map increases, the number of buckets also increases, which can lead to more collisions and slower lookup time. Therefore, the time complexity for finding a value in a HashMap is O(1) on average, but it can be O(n) in the worst case scenario.Rate this question:
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- 36.
Given: 12. import java.util.*; 13. public class Explorer2 { 14. public static void main(String[] args) { 15. TreeSet<Integer> s = new TreeSet<Integer>(); 16. TreeSet<Integer> subs = new TreeSet<Integer>(); 17. for(int i = 606; i < 613; i++) 18. if(i%2 == 0) s.add(i); 19. subs = (TreeSet)s.subSet(608, true, 611, true); 20. s.add(629); 21. System.out.println(s + " " + subs); 22. } 23. } What is the result?
- A.
Compilation fails.
- B.
An exception is thrown at runtime.
- C.
[608, 610, 612, 629] [608, 610]
- D.
[608, 610, 612, 629] [608, 610, 629]
- E.
[606, 608, 610, 612, 629] [608, 610]
- F.
[606, 608, 610, 612, 629] [608, 610, 629]
Correct Answer
E. [606, 608, 610, 612, 629] [608, 610]Explanation
The code creates a TreeSet named "s" and adds even numbers between 606 and 612 to it. Then, it creates a subset of "s" called "subs" using the subSet() method, with the range [608, 611] inclusive. After that, it adds the number 629 to "s". The output will be [606, 608, 610, 612, 629] for "s" and [608, 610] for "subs".Rate this question:
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- 37.
Given: 1. public class Drink implements Comparable { 2. public String name; 3. public int compareTo(Object o) { 4. return 0; 5. } 6. } and: 20. Drink one = new Drink(); 21. Drink two = new Drink(); 22. one.name= "Coffee"; 23. two.name= "Tea"; 24. TreeSet set = new TreeSet(); 25. set.add(one); 26. set.add(two); A programmer iterates over the TreeSet and prints the name of each Drink object. What is the result?
- A.
Tea
- B.
Coffee
- C.
Coffee Tea
- D.
Compilation fails.
- E.
The code runs with no output.
- F.
An exception is thrown at runtime.
Correct Answer
B. CoffeeExplanation
The code will iterate over the TreeSet and print the name of each Drink object. Since the compareTo method in the Drink class returns 0, the TreeSet will consider both Drink objects as equal and will only store one of them. When iterating over the TreeSet, it will only print the name of the Drink object that was stored, which is "Coffee". Therefore, the result will be "Coffee".Rate this question:
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- 38.
A programmer must create a generic class MinMax and the type parameter of MinMax must implement Comparable. Which implementation of MinMax will compile?
- A.
Class MinMax { E min = null; E max = null; public MinMax() {} public void put(E value) { /* store min or max */ }
- B.
Class MinMax { E min = null; E max = null; public MinMax() {} public void put(E value) { /* store min or max */ }
- C.
Class MinMax { E min = null; E max = null; public MinMax() {} public void put(E value) { /* store min or max */ }
- D.
Class MinMax { E min = null; E max = null; public MinMax() {} public void put(E value) { /* store min or max */ }
Correct Answer
A. Class MinMax { E min = null; E max = null; public MinMax() {} public void put(E value) { /* store min or max */ }Explanation
The correct implementation of the MinMax class is the first one provided. This is because it correctly declares the type parameter E and ensures that it implements the Comparable interface. The put() method allows for storing either the minimum or maximum value based on the implementation.Rate this question:
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- 39.
Given: 1. import java.util.*; 2. public class Example { 3. public static void main(String[] args) { 4. // insert code here 5. set.add(new Integer(2)); 6. set.add(new Integer(1)); 7. System.out.println(set); 8. } 9. } Which code, inserted at line 4, guarantees that this program will output [1, 2]?
- A.
Set set = new TreeSet();
- B.
Set set = new HashSet();
- C.
Set set = new SortedSet();
- D.
List set = new SortedList();
- E.
Set set = new LinkedHashSet();
Correct Answer
A. Set set = new TreeSet();Explanation
The correct answer is "Set set = new TreeSet();". This code guarantees that the program will output [1, 2] because TreeSet is an implementation of the Set interface that stores elements in sorted order. When elements are added to a TreeSet, they are automatically sorted in ascending order. Therefore, when the integers 2 and 1 are added to the TreeSet, they will be stored in sorted order, resulting in the output [1, 2].Rate this question:
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- 40.
Given: 5. class A { 6. void foo() throws Exception { throw new Exception(); } 7. } 8. class SubB2 extends A { 9. void foo() { System.out.println("B "); } 10. } 11. class Tester { 12. public static void main(String[] args) { 13. A a = new SubB2(); 14. a.foo(); 15. } 16. } What is the result?
- A.
B
- B.
B, followed by an Exception.
- C.
Compilation fails due to an error on line 9.
- D.
Compilation fails due to an error on line 14.
- E.
An Exception is thrown with no other output.
Correct Answer
D. Compilation fails due to an error on line 14.Explanation
The code fails to compile on line 14 because the method foo() in class A throws an exception, and when overriding a method, the subclass must not throw a broader exception or a new checked exception. In this case, the subclass SubB2 does not declare that it throws an exception, causing a compilation error.Rate this question:
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- 41.
Given: 84. try { 85. ResourceConnection con = resourceFactory.getConnection(); 86. Results r = con.query("GET INFO FROM CUSTOMER"); 87. info = r.getData(); 88. con.close(); 89. } catch (ResourceException re) { 90. errorLog.write(re.getMessage()); 91. } 92. return info; Which statement is true if a ResourceException is thrown on line 86?
- A.
Line 92 will not execute.
- B.
The connection will not be retrieved in line 85.
- C.
The resource connection will not be closed on line 88.
- D.
The enclosing method will throw an exception to its caller.
Correct Answer
C. The resource connection will not be closed on line 88.Explanation
If a ResourceException is thrown on line 86, it means that there was an error while querying the resource. In this case, the code in the catch block on line 90 will be executed, which writes the error message to the error log. However, since the exception occurred before reaching line 88, the resource connection will not be closed. Therefore, the statement "The resource connection will not be closed on line 88" is true.Rate this question:
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- 42.
Given: 3. public class Breaker { 4. static String o = ""; 5. public static void main(String[] args) { 6. z: 7. o = o + 2; 8. for(int x = 3; x < 8; x++) { 9. if(x==4) break; 10. if(x==6) break z; 11. o = o + x; 12. } 13. System.out.println(o); 14. } 15. } What is the result?
- A.
23
- B.
234
- C.
235
- D.
2345
- E.
2357
- F.
23457
- G.
Compilation fails.
Correct Answer
G. Compilation fails.Explanation
The code will fail to compile because the label "z" on line 6 is not used anywhere else in the code. The break statement on line 10 references the label "z", but since it is not used, the code will not compile.Rate this question:
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- 43.
Given: 11. public void go(int x) { 12. assert (x > 0); 13. switch(x) { 14. case 2: ; 15. default: assert false; 16. } 17. } 18. private void go2(int x) { assert (x < 0); } Which statement is true?
- A.
All of the assert statements are used appropriately.
- B.
Only the assert statement on line 12 is used appropriately.
- C.
Only the assert statement on line 15 is used appropriately.
- D.
Only the assert statement on line 18 is used appropriately.
- E.
Only the assert statements on lines 12 and 15 are used appropriately.
- F.
Only the assert statements on lines 12 and 18 are used appropriately.
- G.
Only the assert statements on lines 15 and 18 are used appropriately.
Correct Answer
G. Only the assert statements on lines 15 and 18 are used appropriately.Explanation
The correct answer is "Only the assert statements on lines 15 and 18 are used appropriately." This is because in line 15, the assert statement is used to ensure that the default case in the switch statement is never reached. In line 18, the assert statement is used to check that the value of x is less than 0. Both of these assert statements are used correctly to enforce certain conditions in the code.Rate this question:
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- 44.
Given: 11. public static void main(String[] args) { 12. try { 13. args = null; 14. args[0] = "test"; 15. System.out.println(args[0]); 16. } catch (Exception ex) { 17. System.out.println("Exception"); 18. } catch (NullPointerException npe) { 19. System.out.println("NullPointerException"); 20. } 21. } What is the result?
- A.
Test
- B.
Exception
- C.
Compilation fails.
- D.
NullPointerException
Correct Answer
C. Compilation fails.Explanation
The code will fail to compile because on line 14, the program is trying to assign a value to the first element of the "args" array, which has been set to null on line 13. Since the "args" array is null, it cannot be accessed and therefore the code will not compile.Rate this question:
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- 45.
Given: 11. public static void main(String[] args) { 12. for (int i = 0; i <= 10; i++) { 13. if (i > 6) break; 14. } 15. System.out.println(i); 16. } What is the result?
- A.
6
- B.
7
- C.
10
- D.
11
- E.
Compilation fails.
- F.
An exception is thrown at runtime.
Correct Answer
E. Compilation fails.Explanation
The code will fail to compile because the variable "i" is declared within the for loop and cannot be accessed outside of it. Therefore, the statement on line 15, which tries to print the value of "i", will result in a compilation error.Rate this question:
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- 46.
Given: 11. class X { public void foo() { System.out.print("X "); } } 12. 13. public class SubB extends X { 14. public void foo() throws RuntimeException { 15. super.foo(); 16. if (true) throw new RuntimeException(); 17. System.out.print("B "); 18. } 19. public static void main(String[] args) { 20. new SubB().foo(); 21. } 22. } What is the result?
- A.
X, followed by an Exception.
- B.
No output, and an Exception is thrown.
- C.
Compilation fails due to an error on line 14.
- D.
Compilation fails due to an error on line 16.
- E.
Compilation fails due to an error on line 17.
- F.
X, followed by an Exception, followed by B.
Correct Answer
A. X, followed by an Exception.Explanation
The correct answer is X, followed by an Exception. This is because the foo() method in the SubB class overrides the foo() method in the X class. In the overridden method, it first calls the foo() method of the superclass using super.foo(). Then, it throws a RuntimeException. Since there is no try-catch block to handle the exception, it is propagated up the call stack and the program terminates after printing "X".Rate this question:
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- 47.
Given: 11. public void testIfA() { 12. if (testIfB("True")) { 13. System.out.println("True"); 14. } else { 15. System.out.println("Not true"); 16. } 17. } 18. public Boolean testIfB(String str) { 19. return Boolean.valueOf(str); 20. } What is the result when method testIfA is invoked?
- A.
True
- B.
Not true
- C.
An exception is thrown at runtime.
- D.
Compilation fails because of an error at line 12.
- E.
Compilation fails because of an error at line 19.
Correct Answer
A. TrueExplanation
The result when method testIfA is invoked is "True". This is because the testIfB method is called with the argument "True", which returns a Boolean value of true. Since the if statement evaluates this Boolean value as true, the code block in line 13 is executed, which prints "True" to the console.Rate this question:
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- 48.
Which can appropriately be thrown by a programmer using Java SE technology to create a desktop application?
- A.
ClassCastException
- B.
NullPointerException
- C.
NoClassDefFoundError
- D.
NumberFormatException
- E.
ArrayIndexOutOfBoundsException
Correct Answer
D. NumberFormatExceptionExplanation
The NumberFormatException can be thrown by a programmer using Java SE technology to create a desktop application when there is an attempt to convert a string into a numeric type, but the string does not have the appropriate format. This exception is usually thrown when using methods like parseInt() or parseDouble() to convert strings into integers or doubles, respectively. If the string cannot be parsed into a valid number format, the NumberFormatException is thrown.Rate this question:
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- 49.
Which two code fragments are most likely to cause a StackOverflowError? (Choose two.)
- A.
Int []x = {1,2,3,4,5}; for(int y = 0; y < 6; y++) System.out.println(x[y]);
- B.
Static int[] x = {7,6,5,4}; static { x[1] = 8; x[4] = 3; }
- C.
for(int y = 10; y < 10; y++) doStuff(y);
- D.
Void doOne(int x) { doTwo(x); } void doTwo(int y) { doThree(y); } void doThree(int z) { doTwo(z); }
- E.
For(int x = 0; x < 1000000000; x++) doStuff(x);
- F.
Void counter(int i) { counter(++i); }
Correct Answer(s)
D. Void doOne(int x) { doTwo(x); } void doTwo(int y) { doThree(y); } void doThree(int z) { doTwo(z); }
F. Void counter(int i) { counter(++i); }Explanation
The first code fragment is likely to cause a StackOverflowError because it creates an array of size 5 and then tries to access index 5, which is out of bounds. This results in an infinite loop of trying to access the same index, causing the stack to overflow.
The second code fragment is also likely to cause a StackOverflowError because it creates a recursive function where doThree calls doTwo and doTwo calls doThree. This creates an infinite loop of function calls, causing the stack to overflow.
Both of these scenarios involve infinite loops or recursive function calls that do not have a base case or termination condition, leading to a stack overflow.Rate this question:
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- 50.
Given: 11. public static void main(String[] args) { 12. Integer i = new Integer(1) + new Integer(2); 13. switch(i) { 14. case 3: System.out.println("three"); break; 15. default: System.out.println("other"); break; 16. } 17. } What is the result?
- A.
Three
- B.
Other
- C.
An exception is thrown at runtime.
- D.
Compilation fails because of an error on line 12.
- E.
Compilation fails because of an error on line 13.
- F.
Compilation fails because of an error on line 15.
Correct Answer
A. ThreeExplanation
The code will output "three". This is because on line 12, the code is creating two Integer objects, adding them together using the "+" operator, and assigning the result to the variable "i". This is possible because of autoboxing, which automatically converts the primitive int values to Integer objects. The value of "i" is then used in the switch statement on line 13. Since the value of "i" is 3, the case 3 is matched and the code prints "three".Rate this question:
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Mar 20, 2023Quiz Edited by
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