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- Question
Which of the following statements about the hashcode() method are incorrect?
- The value returned by hashcode() is used in some collection classes to help locate objects.
- The hashcode() method is required to return a positive int value.
- The hashcode() method in the String class is the one inherited from Object.
- Two new empty String objects will produce identical hashcodes.
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 1 and 4
- Correct Answer
- 2 and 3
Explanation(2) is an incorrect statement because there is no such requirement.(3) is an incorrect statement and therefore a correct answer because the hashcode for a string is computed from the characters in the string.
More questions
- 1. 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.- 2. 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.- 3. 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).
- 4. 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.
- 5. 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.
- 6. 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.
- 7. 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} };
- 8. 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".- 9. Which will legally declare, construct, and initialize an array?
Options- A. int [] myList = {"1", "2", "3"};
- B. int [] myList = (5, 8, 2);
- C. int myList [] [] = {4,9,7,0};
- D. int myList [] = {4, 3, 7}; Discuss
Correct Answer: int myList [] = {4, 3, 7};
Explanation:
The only legal array declaration and assignment statement is Option DOption A is wrong because it initializes an int array with String literals.
Option B is wrong because it use something other than curly braces for the initialization.
Option C is wrong because it provides initial values for only one dimension, although the declared array is a two-dimensional array.
- 10. What will be the output of the program?
class BitShift { public static void main(String [] args) { int x = 0x80000000; System.out.print(x + " and "); x = x >>> 31; System.out.println(x); } }
Options- A. -2147483648 and 1
- B. 0x80000000 and 0x00000001
- C. -2147483648 and -1
- D. 1 and -2147483648 Discuss
Correct Answer: -2147483648 and 1
Explanation:
Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
Option B is incorrect because the output method print() always displays integers in base 10.
Option D is incorrect because this is the reverse order of the two output numbers.
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- 1. What will be the output of the program?