Discussion
Home ‣ Java Programming ‣ Declarations and Access Control See What Others Are Saying!
- Question
What will be the output of the program?
public class A { void A() /* Line 3 */ { System.out.println("Class A"); } public static void main(String[] args) { new A(); } }
Options- A. Class A
- B. Compilation fails.
- C. An exception is thrown at line 3.
- D. The code executes with no output.
- Correct Answer
- The code executes with no output.
ExplanationOption D is correct. The specification at line 3 is for a method and not a constructor and this method is never called therefore there is no output. The constructor that is called is the default constructor.
More questions
- 1. What will be the output of the program?
public class CommandArgsTwo { public static void main(String [] argh) { int x; x = argh.length; for (int y = 1; y <= x; y++) { System.out.print(" " + argh[y]); } } }and the command-line invocation is> java CommandArgsTwo 1 2 3
Options- A. 0 1 2
- B. 1 2 3
- C. 0 0 0
- D. An exception is thrown at runtime Discuss
Correct Answer: An exception is thrown at runtime
Explanation:
An exception is thrown because at some point in (System.out.print(" " + argh[y]);), the value of x will be equal to y, resulting in an attempt to access an index out of bounds for the array. Remember that you can access only as far as length - 1, so loop logical tests should use x < someArray.length as opposed to x < = someArray.length.- 2. What will be the output of the program?
int i = l, j = -1; switch (i) { case 0, 1: j = 1; /* Line 4 */ case 2: j = 2; default: j = 0; } System.out.println("j = " + j);
Options- A. j = -1
- B. j = 0
- C. j = 1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
The case statement takes only a single argument. The case statement on line 4 is given two arguments so the compiler complains.- 3. What will be the output of the program?
public class BoolTest { public static void main(String [] args) { int result = 0; Boolean b1 = new Boolean("TRUE"); Boolean b2 = new Boolean("true"); Boolean b3 = new Boolean("tRuE"); Boolean b4 = new Boolean("false"); if (b1 == b2) /* Line 10 */ result = 1; if (b1.equals(b2) ) /* Line 12 */ result = result + 10; if (b2 == b4) /* Line 14 */ result = result + 100; if (b2.equals(b4) ) /* Line 16 */ result = result + 1000; if (b2.equals(b3) ) /* Line 18 */ result = result + 10000; System.out.println("result = " + result); } }
Options- A. 0
- B. 1
- C. 10
- D. 10010 Discuss
Correct Answer: 10010
Explanation:
Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.- 4. What will be the output of the program?
int i = 0, j = 5; tp: for (;;) { i++; for (;;) { if(i > --j) { break tp; } } System.out.println("i =" + i + ", j = " + j);
Options- A. i = 1, j = 0
- B. i = 1, j = 4
- C. i = 3, j = 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
If you examine the code carefully you will notice a missing curly bracket at the end of the code, this would cause the code to fail.- 5. Which two statements are equivalent?
- 3/2
- 3<2
- 3*4
- 3<<2
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 4 Discuss
Correct Answer: 3 and 4
Explanation:
(1) is wrong. 3/2 = 1 (integer arithmetic).(2) is wrong. 3 < 2 = false.
(3) is correct. 3 * 4 = 12.
(4) is correct. 3 <<2= 12. In binary 3 is 11, now shift the bits two places to the left and we get 1100 which is 12 in binary (3*2*2).
- 6. Which three guarantee that a thread will leave the running state?
- yield()
- wait()
- notify()
- notifyAll()
- sleep(1000)
- aLiveThread.join()
- Thread.killThread()
Options- A. 1, 2 and 4
- B. 2, 5 and 6
- C. 3, 4 and 7
- D. 4, 5 and 7 Discuss
Correct Answer: 2, 5 and 6
Explanation:
(2) is correct because wait() always causes the current thread to go into the object's wait pool.(5) is correct because sleep() will always pause the currently running thread for at least the duration specified in the sleep argument (unless an interrupted exception is thrown).
(6) is correct because, assuming that the thread you're calling join() on is alive, the thread calling join() will immediately block until the thread you're calling join() on is no longer alive.
(1) is wrong, but tempting. The yield() method is not guaranteed to cause a thread to leave the running state, although if there are runnable threads of the same priority as the currently running thread, then the current thread will probably leave the running state.
(3) and (4) are incorrect because they don't cause the thread invoking them to leave the running state.
(7) is wrong because there's no such method.
- 7. What will be the output of the program?
int i = 1, j = 10; do { if(i > j) { break; } j--; } while (++i < 5); System.out.println("i = " + i + " and j = " + j);
Options- A. i = 6 and j = 5
- B. i = 5 and j = 5
- C. i = 6 and j = 4
- D. i = 5 and j = 6 Discuss
Correct Answer: i = 5 and j = 6
Explanation:
This loop is a do-while loop, which always executes the code block within the block at least once, due to the testing condition being at the end of the loop, rather than at the beginning. This particular loop is exited prematurely if i becomes greater than j.The order is, test i against j, if bigger, it breaks from the loop, decrements j by one, and then tests the loop condition, where a pre-incremented by one i is tested for being lower than 5. The test is at the end of the loop, so i can reach the value of 5 before it fails. So it goes, start:
1, 10
2, 9
3, 8
4, 7
5, 6 loop condition fails.
- 8. What will be the output of the program?
public class Q126 implements Runnable { private int x; private int y; public static void main(String [] args) { Q126 that = new Q126(); (new Thread(that)).start( ); /* Line 8 */ (new Thread(that)).start( ); /* Line 9 */ } public synchronized void run( ) /* Line 11 */ { for (;;) /* Line 13 */ { x++; y++; System.out.println("x = " + x + "y = " + y); } } }
Options- A. An error at line 11 causes compilation to fail
- B. Errors at lines 8 and 9 cause compilation to fail.
- C. The program prints pairs of values for x and y that might not always be the same on the same line (for example, "x=2, y=1")
- D. The program prints pairs of values for x and y that are always the same on the same line (for example, "x=1, y=1". In addition, each value appears once (for example, "x=1, y=1" followed by "x=2, y=2") Discuss
Correct Answer: The program prints pairs of values for x and y that are always the same on the same line (for example, "x=1, y=1". In addition, each value appears once (for example, "x=1, y=1" followed by "x=2, y=2")
Explanation:
The synchronized code is the key to answering this question. Because x and y are both incremented inside the synchronized method they are always incremented together. Also keep in mind that the two threads share the same reference to the Q126 object.Also note that because of the infinite loop at line 13, only one thread ever gets to execute.
- 9. Which cause a compiler error?
Options- A. int[ ] scores = {3, 5, 7};
- B. int [ ][ ] scores = {2,7,6}, {9,3,45};
- C. String cats[ ] = {"Fluffy", "Spot", "Zeus"};
- D. boolean results[ ] = new boolean [] {true, false, true};
- E. Integer results[ ] = {new Integer(3), new Integer(5), new Integer(8)}; Discuss
Correct Answer: int [ ][ ] scores = {2,7,6}, {9,3,45};
Explanation:
Option B generates a compiler error: <identifier> expected. The compiler thinks you are trying to create two arrays because there are two array initialisers to the right of the equals, whereas your intention was to create one 3 x 3 two-dimensional array.To correct the problem and make option B compile you need to add an extra pair of curly brackets:
int [ ] [ ] scores = { {2,7,6}, {9,3,45} };
- 10. What will be the output of the program?
class SSBool { public static void main(String [] args) { boolean b1 = true; boolean b2 = false; boolean b3 = true; if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */ System.out.print("ok "); if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/ System.out.println("dokey"); } }
Options- A. ok
- B. dokey
- C. ok dokey
- D. No output is produced
- E. Compilation error Discuss
Correct Answer: dokey
Explanation:
The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. What will be the output of the program?