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- Question
You want subclasses in any package to have access to members of a superclass. Which is the most restrictive access that accomplishes this objective?
Options- A. public
- B. private
- C. protected
- D. transient
- Correct Answer
- protected
ExplanationAccess modifiers dictate which classes, not which instances, may access features.Methods and variables are collectively known as members. Method and variable members are given access control in exactly the same way.
private makes a member accessible only from within its own class
protected makes a member accessible only to classes in the same package or subclass of the class
default access is very similar to protected (make sure you spot the difference) default access makes a member accessible only to classes in the same package.
public means that all other classes regardless of the package that they belong to, can access the member (assuming the class itself is visible)
final makes it impossible to extend a class, when applied to a method it prevents a method from being overridden in a subclass, when applied to a variable it makes it impossible to reinitialise a variable once it has been initialised
abstract declares a method that has not been implemented.
transient indicates that a variable is not part of the persistent state of an object.
volatile indicates that a thread must reconcile its working copy of the field with the master copy every time it accesses the variable.
After examining the above it should be obvious that the access modifier that provides the most restrictions for methods to be accessed from the subclasses of the class from another package is C - protected. A is also a contender but C is more restrictive, B would be the answer if the constraint was the "same package" instead of "any package" in other words the subclasses clause in the question eliminates default.
More questions
- 1. Which is a reserved word in the Java programming language?
Options- A. method
- B. native
- C. subclasses
- D. reference
- E. array Discuss
Correct Answer: native
Explanation:
The word "native" is a valid keyword, used to modify a method declaration.Option A, D and E are not keywords. Option C is wrong because the keyword for subclassing in Java is extends, not 'subclasses'.
- 2. Which statement is true for the class java.util.ArrayList?
Options- A. The elements in the collection are ordered.
- B. The collection is guaranteed to be immutable.
- C. The elements in the collection are guaranteed to be unique.
- D. The elements in the collection are accessed using a unique key. Discuss
Correct Answer: The elements in the collection are ordered.
Explanation:
Yes, always the elements in the collection are ordered.- 3. What will be the output of the program?
public class Test { public static void main(String[] args) { final StringBuffer a = new StringBuffer(); final StringBuffer b = new StringBuffer(); new Thread() { public void run() { System.out.print(a.append("A")); synchronized(b) { System.out.print(b.append("B")); } } }.start(); new Thread() { public void run() { System.out.print(b.append("C")); synchronized(a) { System.out.print(a.append("D")); } } }.start(); } }
Options- A. ACCBAD
- B. ABBCAD
- C. CDDACB
- D. Indeterminate output Discuss
Correct Answer: Indeterminate output
Explanation:
It gives different output while executing the same compiled code at different times.- 4. Which is true about a method-local inner class?
Options- A. It must be marked final.
- B. It can be marked abstract.
- C. It can be marked public.
- D. It can be marked static. Discuss
Correct Answer: It can be marked abstract.
Explanation:
Option B is correct because a method-local inner class can be abstract, although it means a subclass of the inner class must be created if the abstract class is to be used (so an abstract method-local inner class is probably not useful).Option A is incorrect because a method-local inner class does not have to be declared final (although it is legal to do so).
C and D are incorrect because a method-local inner class cannot be made public (remember-you cannot mark any local variables as public), or static.
- 5. Given that all methods of class FileOutputStream, including close(), throw an IOException, which of these is true?
import java.io.*; public class MyProgram { public static void main(String args[]) { FileOutputStream out = null; try { out = new FileOutputStream("test.txt"); out.write(122); } catch(IOException io) { System.out.println("IO Error."); } finally { out.close(); } } }
Options- A. This program will compile successfully.
- B. This program fails to compile due to an error at line 4.
- C. This program fails to compile due to an error at line 6.
- D. This program fails to compile due to an error at line 18. Discuss
Correct Answer: This program fails to compile due to an error at line 18.
Explanation:
Any method (in this case, the main() method) that throws a checked exception (in this case, out.close() ) must be called within a try clause, or the method must declare that it throws the exception. Either main() must declare that it throws an exception, or the call to out.close() in the finally block must fall inside a (in this case nested) try-catch block.- 6. What will be the output of the program?
public class Test { public static void main(String [] args) { int I = 1; do while ( I < 1 ) System.out.print("I is " + I); while ( I > 1 ) ; } }
Options- A. I is 1
- B. I is 1 I is 1
- C. No output is produced.
- D. Compilation error Discuss
Correct Answer: No output is produced.
Explanation:
There are two different looping constructs in this problem. The first is a do-while loop and the second is a while loop, nested inside the do-while. The body of the do-while is only a single statement-brackets are not needed. You are assured that the while expression will be evaluated at least once, followed by an evaluation of the do-while expression. Both expressions are false and no output is produced.- 7. What will be the output of the program?
class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); Thread x = new Thread(t); x.start(); /* Line 7 */ } public void run() { for(int i = 0; i < 3; ++i) { System.out.print(i + ".."); } } }
Options- A. Compilation fails.
- B. 1..2..3..
- C. 0..1..2..3..
- D. 0..1..2.. Discuss
Correct Answer: 0..1..2..
Explanation:
The thread MyThread will start and loop three times (from 0 to 2).Option A is incorrect because the Thread class implements the Runnable interface; therefore, in line 7, Thread can take an object of type Thread as an argument in the constructor.
Option B and C are incorrect because the variable i in the for loop starts with a value of 0 and ends with a value of 2.
- 8. Which of the following is/are legal method declarations?
- protected abstract void m1();
- static final void m1(){}
- synchronized public final void m1() {}
- private native void m1();
Options- A. 1 and 3
- B. 2 and 4
- C. 1 only
- D. All of them are legal declarations. Discuss
Correct Answer: All of them are legal declarations.
Explanation:
All the given statements are legal declarations.- 9. Which three statements are true?
- f1 == f2
- f1 == f3
- f2 == f1[1]
- x == f1[0]
- f == f1[0]
import java.awt.Button; class CompareReference { public static void main(String [] args) { float f = 42.0f; float [] f1 = new float[2]; float [] f2 = new float[2]; float [] f3 = f1; long x = 42; f1[0] = 42.0f; } }
Options- A. 1, 2 and 3
- B. 2, 4 and 5
- C. 3, 4 and 5
- D. 1, 4 and 5 Discuss
Correct Answer: 2, 4 and 5
Explanation:
(2) is correct because the reference variables f1 and f3 refer to the same array object.(4) is correct because it is legal to compare integer and floating-point types.
(5) is correct because it is legal to compare a variable with an array element.
(3) is incorrect because f2 is an array object and f1[1] is an array element.
- 10. Which one create an anonymous inner class from within class Bar?
class Boo { Boo(String s) { } Boo() { } } class Bar extends Boo { Bar() { } Bar(String s) {super(s);} void zoo() { // insert code here } }
Options- A. Boo f = new Boo(24) { };
- B. Boo f = new Bar() { };
- C. Bar f = new Boo(String s) { };
- D. Boo f = new Boo.Bar(String s) { }; Discuss
Correct Answer: Boo f = new Bar() { };
Explanation:
Option B is correct because anonymous inner classes are no different from any other class when it comes to polymorphism. That means you are always allowed to declare a reference variable of the superclass type and have that reference variable refer to an instance of a subclass type, which in this case is an anonymous subclass of Bar. Since Bar is a subclass of Boo, it all works.Option A is incorrect because it passes an int to the Boo constructor, and there is no matching constructor in the Boo class.
Option C is incorrect because it violates the rules of polymorphism—you cannot refer to a superclass type using a reference variable declared as the subclass type. The superclass is not guaranteed to have everything the subclass has.
Option D uses incorrect syntax.
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