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- Question
What will be the output of the program?
class MyThread extends Thread { MyThread() { System.out.print(" MyThread"); } public void run() { System.out.print(" bar"); } public void run(String s) { System.out.println(" baz"); } } public class TestThreads { public static void main (String [] args) { Thread t = new MyThread() { public void run() { System.out.println(" foo"); } }; t.start(); } }
Options- A. foo
- B. MyThread foo
- C. MyThread bar
- D. foo bar
- Correct Answer
- MyThread foo
ExplanationOption B is correct because in the first line of main we're constructing an instance of an anonymous inner class extending from MyThread. So the MyThread constructor runs and prints "MyThread". The next statement in main invokes start() on the new thread instance, which causes the overridden run() method (the run() method defined in the anonymous inner class) to be invoked, which prints "foo"
More questions
- 1. 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.- 2. 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.
- 3. 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.
- 4. 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.- 5. 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.- 6. 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()
- 7. 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.
- 8. What is the most restrictive access modifier that will allow members of one class to have access to members of another class in the same package?
Options- A. public
- B. abstract
- C. protected
- D. synchronized
- E. default access Discuss
Correct Answer: default access
Explanation:
default access is the "package oriented" access modifier.Option A and C are wrong because public and protected are less restrictive. Option B and D are wrong because abstract and synchronized are not access modifiers.
- 9. Which of the following statements is true?
Options- A. If assertions are compiled into a source file, and if no flags are included at runtime, assertions will execute by default.
- B. As of Java version 1.4, assertion statements are compiled by default.
- C. With the proper use of runtime arguments, it is possible to instruct the VM to disable assertions for a certain class, and to enable assertions for a certain package, at the same time.
- D. When evaluating command-line arguments, the VM gives -ea flags precedence over -da flags. Discuss
Correct Answer: With the proper use of runtime arguments, it is possible to instruct the VM to disable assertions for a certain class, and to enable assertions for a certain package, at the same time.
Explanation:
Option C is true because multiple VM flags can be used on a single invocation of a Java program.Option A is incorrect because at runtime assertions are ignored by default.
Option B is incorrect because as of Java 1.4 you must add the argument -source 1.4 to the command line if you want the compiler to compile assertion statements.
Option D is incorrect because the VM evaluates all assertion flags left to right.
- 10. What will be the output of the program?
public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. 2 1 0 1 0 0
- D. 2 1 2 0 1 2 Discuss
Correct Answer: 2 1 2 0 1 2
Explanation:
The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.
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- 1. What will be the output of the program?