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- Question
What will be the output of the program?
class Super { public Integer getLength() { return new Integer(4); } } public class Sub extends Super { public Long getLength() { return new Long(5); } public static void main(String[] args) { Super sooper = new Super(); Sub sub = new Sub(); System.out.println( sooper.getLength().toString() + "," + sub.getLength().toString() ); } }
Options- A. 4, 4
- B. 4, 5
- C. 5, 4
- D. Compilation fails.
- Correct Answer
- Compilation fails.
ExplanationOption D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.
More questions
- 1. Which cannot directly cause a thread to stop executing?
Options- A. Calling the SetPriority() method on a Thread object.
- B. Calling the wait() method on an object.
- C. Calling notify() method on an object.
- D. Calling read() method on an InputStream object. Discuss
Correct Answer: Calling notify() method on an object.
Explanation:
Option C is correct. notify() - wakes up a single thread that is waiting on this object's monitor.- 2. What will be the output of the program?
public class Foo { public static void main(String[] args) { try { return; } finally { System.out.println( "Finally" ); } } }
Options- A. Finally
- B. Compilation fails.
- C. The code runs with no output.
- D. An exception is thrown at runtime. Discuss
- An exception arising in the finally block itself.
- The death of the thread.
- The use of System.exit()
- Turning off the power to the CPU.
Correct Answer: Finally
Explanation:
If you put a finally block after a try and its associated catch blocks, then once execution enters the try block, the code in that finally block will definitely be executed except in the following circumstances:- 3. What will be the output of the program?
public class Test138 { public static void stringReplace (String text) { text = text.replace ('j' , 'c'); /* Line 5 */ } public static void bufferReplace (StringBuffer text) { text = text.append ("c"); /* Line 9 */ } public static void main (String args[]) { String textString = new String ("java"); StringBuffer textBuffer = new StringBuffer ("java"); /* Line 14 */ stringReplace(textString); bufferReplace(textBuffer); System.out.println (textString + textBuffer); } }
Options- A. java
- B. javac
- C. javajavac
- D. Compile error Discuss
Correct Answer: javajavac
Explanation:
A string is immutable, it cannot be changed, that's the reason for the StringBuffer class. The stringReplace method does not change the string declared on line 14, so this remains set to "java".Method parameters are always passed by value - a copy is passed into the method - if the copy changes, the original remains intact, line 5 changes the reference i.e. text points to a new String object, however this is lost when the method completes. The textBuffer is a StringBuffer so it can be changed.
This change is carried out on line 9, so "java" becomes "javac", the text reference on line 9 remains unchanged. This gives us the output of "javajavac"
- 4. What will be the output of the program?
public class WrapTest { public static void main(String [] args) { int result = 0; short s = 42; Long x = new Long("42"); Long y = new Long(42); Short z = new Short("42"); Short x2 = new Short(s); Integer y2 = new Integer("42"); Integer z2 = new Integer(42); if (x == y) /* Line 13 */ result = 1; if (x.equals(y) ) /* Line 15 */ result = result + 10; if (x.equals(z) ) /* Line 17 */ result = result + 100; if (x.equals(x2) ) /* Line 19 */ result = result + 1000; if (x.equals(z2) ) /* Line 21 */ result = result + 10000; System.out.println("result = " + result); } }
Options- A. result = 1
- B. result = 10
- C. result = 11
- D. result = 11010 Discuss
Correct Answer: result = 10
Explanation:
Line 13 fails because == compares reference values, not object values. Line 15 succeeds because both String and primitive wrapper constructors resolve to the same value (except for the Character wrapper). Lines 17, 19, and 21 fail because the equals() method fails if the object classes being compared are different and not in the same tree hierarchy.- 5. Which three are valid declarations of a float?
- float f1 = -343;
- float f2 = 3.14;
- float f3 = 0x12345;
- float f4 = 42e7;
- float f5 = 2001.0D;
- float f6 = 2.81F;
Options- A. 1, 2, 4
- B. 2, 3, 5
- C. 1, 3, 6
- D. 2, 4, 6 Discuss
Correct Answer: 1, 3, 6
Explanation:
(1) and (3) are integer literals (32 bits), and integers can be legally assigned to floats (also 32 bits). (6) is correct because (F) is appended to the literal, declaring it as a float rather than a double (the default for floating point literals).(2), (4),and (5) are all doubles.
- 6. Which statement is true?
Options- A. catch(X x) can catch subclasses of X where X is a subclass of Exception.
- B. The Error class is a RuntimeException.
- C. Any statement that can throw an Error must be enclosed in a try block.
- D. Any statement that can throw an Exception must be enclosed in a try block. Discuss
Correct Answer: catch(X x) can catch subclasses of X where X is a subclass of Exception.
Explanation:
Option A is correct. If the class specified in the catch clause does have subclasses, any exception object that subclasses the specified class will be caught as well.Option B is wrong. The error class is a subclass of Throwable and not Runtime Exception.
Option C is wrong. You do not catch this class of error.
Option D is wrong. An exception can be thrown to the next method higher up the call stack.
- 7. What will be the output of the program?
import java.util.*; public class NewTreeSet2 extends NewTreeSet { public static void main(String [] args) { NewTreeSet2 t = new NewTreeSet2(); t.count(); } } protected class NewTreeSet { void count() { for (int x = 0; x < 7; x++,x++ ) { System.out.print(" " + x); } } }
Options- A. 0 2 4
- B. 0 2 4 6
- C. Compilation fails at line 2
- D. Compilation fails at line 10 Discuss
Correct Answer: Compilation fails at line 10
Explanation:
Nonnested classes cannot be marked protected (or final for that matter), so the compiler will fail at protected class NewTreeSet.- 8. What will be the output of the program?
int x = 3; int y = 1; if (x = y) /* Line 3 */ { System.out.println("x =" + x); }
Options- A. x = 1
- B. x = 3
- C. Compilation fails.
- D. The code runs with no output. Discuss
Correct Answer: Compilation fails.
Explanation:
Line 3 uses an assignment as opposed to comparison. Because of this, the if statement receives an integer value instead of a boolean. And so the compilation fails.- 9. What will be the output of the program?
public static void main(String[] args) { Object obj = new Object() { public int hashCode() { return 42; } }; System.out.println(obj.hashCode()); }
Options- A. 42
- B. Runtime Exception
- C. Compile Error at line 2
- D. Compile Error at line 5 Discuss
Correct Answer: 42
Explanation:
This code is an example of an anonymous inner class. They can be declared to extend another class or implement a single interface. Since they have no name you can not use the "new" keyword on them.In this case the annoynous class is extending the Object class. Within the {} you place the methods you want for that class. After this class has been declared its methods can be used by that object in the usual way e.g. objectname.annoymousClassMethod()
- 10. What will be the output of the program?
public class SyncTest { public static void main (String [] args) { Thread t = new Thread() { Foo f = new Foo(); public void run() { f.increase(20); } }; t.start(); } } class Foo { private int data = 23; public void increase(int amt) { int x = data; data = x + amt; } }and assuming that data must be protected from corruption, what?if anything?can you add to the preceding code to ensure the integrity of data?
Options- A. Synchronize the run method.
- B. Wrap a synchronize(this) around the call to f.increase().
- C. The existing code will cause a runtime exception.
- D. Synchronize the increase() method Discuss
Correct Answer: Synchronize the increase() method
Explanation:
Option D is correct because synchronizing the code that actually does the increase will protect the code from being accessed by more than one thread at a time.Option A is incorrect because synchronizing the run() method would stop other threads from running the run() method (a bad idea) but still would not prevent other threads with other runnables from accessing the increase() method.
Option B is incorrect for virtually the same reason as A?synchronizing the code that calls the increase() method does not prevent other code from calling the increase() method.
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