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- Question
Which class does not override the equals() and hashCode() methods, inheriting them directly from class Object?
Options- A. java.lang.String
- B. java.lang.Double
- C. java.lang.StringBuffer
- D. java.lang.Character
- Correct Answer
- java.lang.StringBuffer
Explanationjava.lang.StringBuffer is the only class in the list that uses the default methods provided by class Object.
More questions
- 1. Which is a valid keyword in java?
Options- A. interface
- B. string
- C. Float
- D. unsigned Discuss
Correct Answer: interface
Explanation:
interface is a valid keyword.Option B is wrong because although "String" is a class type in Java, "string" is not a keyword.
Option C is wrong because "Float" is a class type. The keyword for the Java primitive is float.
Option D is wrong because "unsigned" is a keyword in C/C++ but not in Java.
- 2. What will be the output of the program?
int i = 1, j = 10; do { if(i++ > --j) /* Line 4 */ { continue; } } while (i < 5); System.out.println("i = " + i + "and j = " + j); /* Line 9 */
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 6
- D. i = 5 and j = 6 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This question is not testing your knowledge of the continue statement. It is testing your knowledge of the order of evaluation of operands. Basically the prefix and postfix unary operators have a higher order of evaluation than the relational operators. So on line 4 the variable i is incremented and the variable j is decremented before the greater than comparison is made. As the loop executes the comparison on line 4 will be:if(i > j)
if(2 > 9)
if(3 > 8)
if(4 > 7)
if(5 > 6) at this point i is not less than 5, therefore the loop terminates and line 9 outputs the values of i and j as 5 and 6 respectively.
The continue statement never gets to execute because i never reaches a value that is greater than j.
- 3. What will be the output of the program?
int i = 0; while(1) { if(i == 4) { break; } ++i; } System.out.println("i = " + i);
Options- A. i = 0
- B. i = 3
- C. i = 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because the argument of the while loop, the condition, must be of primitive type boolean. In Java, 1 does not represent the true state of a boolean, rather it is seen as an integer.- 4. What will be the output of the program?
public class Test { public int aMethod() { static int i = 0; i++; return i; } public static void main(String args[]) { Test test = new Test(); test.aMethod(); int j = test.aMethod(); System.out.println(j); } }
Options- A. 0
- B. 1
- C. 2
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation failed because static was an illegal start of expression - method variables do not have a modifier (they are always considered local).- 5. What will be the output of the program?
int x = l, y = 6; while (y--) { x++; } System.out.println("x = " + x +" y = " + y);
Options- A. x = 6 y = 0
- B. x = 7 y = 0
- C. x = 6 y = -1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because the while loop demands a boolean argument for it's looping condition, but in the code, it's given an int argument.while(true) { //insert code here }
- 6. What will be the output of the program?
class MyThread extends Thread { MyThread() {} MyThread(Runnable r) {super(r); } public void run() { System.out.print("Inside Thread "); } } class MyRunnable implements Runnable { public void run() { System.out.print(" Inside Runnable"); } } class Test { public static void main(String[] args) { new MyThread().start(); new MyThread(new MyRunnable()).start(); } }
Options- A. Prints "Inside Thread Inside Thread"
- B. Prints "Inside Thread Inside Runnable"
- C. Does not compile
- D. Throws exception at runtime Discuss
Correct Answer: Prints "Inside Thread Inside Thread"
Explanation:
If a Runnable object is passed to the Thread constructor, then the run method of the Thread class will invoke the run method of the Runnable object.In this case, however, the run method in the Thread class is overridden by the run method in MyThread class. Therefore the run() method in MyRunnable is never invoked.
Both times, the run() method in MyThread is invoked instead.
- 7. Which statement is true?
Options- A. Assertions can be enabled or disabled on a class-by-class basis.
- B. Conditional compilation is used to allow tested classes to run at full speed.
- C. Assertions are appropriate for checking the validity of arguments in a method.
- D. The programmer can choose to execute a return statement or to throw an exception if an assertion fails. Discuss
Correct Answer: Assertions can be enabled or disabled on a class-by-class basis.
Explanation:
Option A is correct. The assertion status can be set for a named top-level class and any nested classes contained therein. This setting takes precedence over the class loader's default assertion status, and over any applicable per-package default. If the named class is not a top-level class, the change of status will have no effect on the actual assertion status of any class.Option B is wrong. Is there such a thing as conditional compilation in Java?
Option C is wrong. For private methods - yes. But do not use assertions to check the parameters of a public method. An assert is inappropriate in public methods because the method guarantees that it will always enforce the argument checks. A public method must check its arguments whether or not assertions are enabled. Further, the assert construct does not throw an exception of the specified type. It can throw only an AssertionError.
Option D is wrong. Because you're never supposed to handle an assertion failure. That means don't catch it with a catch clause and attempt to recover.
- 8. What will be the output of the program?
class Two { byte x; } class PassO { public static void main(String [] args) { PassO p = new PassO(); p.start(); } void start() { Two t = new Two(); System.out.print(t.x + " "); Two t2 = fix(t); System.out.println(t.x + " " + t2.x); } Two fix(Two tt) { tt.x = 42; return tt; } }
Options- A. null null 42
- B. 0 0 42
- C. 0 42 42
- D. 0 0 0 Discuss
Correct Answer: 0 42 42
Explanation:
In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.- 9. 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 Discuss
Correct Answer: protected
Explanation:
Access 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.
- 10. What will be the output of the program?
public class RTExcept { public static void throwit () { System.out.print("throwit "); throw new RuntimeException(); } public static void main(String [] args) { try { System.out.print("hello "); throwit(); } catch (Exception re ) { System.out.print("caught "); } finally { System.out.print("finally "); } System.out.println("after "); } }
Options- A. hello throwit caught
- B. Compilation fails
- C. hello throwit RuntimeException caught after
- D. hello throwit caught finally after Discuss
Correct Answer: hello throwit caught finally after
Explanation:
The main() method properly catches and handles the RuntimeException in the catch block, finally runs (as it always does), and then the code returns to normal.A, B and C are incorrect based on the program logic described above. Remember that properly handled exceptions do not cause the program to stop executing.
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