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- Question
What two statements are true about properly overridden hashCode() and equals() methods?
- hashCode() doesn't have to be overridden if equals() is.
- equals() doesn't have to be overridden if hashCode() is.
- hashCode() can always return the same value, regardless of the object that invoked it.
- equals() can be true even if it's comparing different objects.
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 3
- Correct Answer
- 3 and 4
Explanation(3) and (4) are correct.(1) and (2) are incorrect because by contract hashCode() and equals() can't be overridden unless both are overridden.
More questions
- 1. Which constructs an anonymous inner class instance?
Options- A. Runnable r = new Runnable() { };
- B. Runnable r = new Runnable(public void run() { });
- C. Runnable r = new Runnable { public void run(){}};
- D. System.out.println(new Runnable() {public void run() { }}); Discuss
Correct Answer: System.out.println(new Runnable() {public void run() { }});
Explanation:
D is correct. It defines an anonymous inner class instance, which also means it creates an instance of that new anonymous class at the same time. The anonymous class is an implementer of the Runnable interface, so it must override the run() method of Runnable.A is incorrect because it doesn't override the run() method, so it violates the rules of interface implementation.
B and C use incorrect syntax.
- 2. What is the name of the method used to start a thread execution?
Options- A. init();
- B. start();
- C. run();
- D. resume(); Discuss
Correct Answer: start();
Explanation:
Option B is Correct. The start() method causes this thread to begin execution; the Java Virtual Machine calls the run method of this thread.Option A is wrong. There is no init() method in the Thread class.
Option C is wrong. The run() method of a thread is like the main() method to an application. Starting the thread causes the object's run method to be called in that separately executing thread.
Option D is wrong. The resume() method is deprecated. It resumes a suspended thread.
- 3. 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.- 4. 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.- 5. 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.- 6. 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.- 7. 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).
- 8. 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.
- 9. 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.
- 10. 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.
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