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- Question
What will be the output of the program?
public class Test { public static void main(String args[]) { int i = 1, j = 0; switch(i) { case 2: j += 6; case 4: j += 1; default: j += 2; case 0: j += 4; } System.out.println("j = " + j); } }
Options- A. j = 0
- B. j = 2
- C. j = 4
- D. j = 6
- Correct Answer
- j = 6
ExplanationBecause there are no break statements, the program gets to the default case and adds 2 to j, then goes to case 0 and adds 4 to the new j. The result is j = 6.
More questions
- 1. 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".- 2. 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.
- 3. 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.
- 4. Which of the following are Java reserved words?
- run
- import
- default
- implement
Options- A. 1 and 2
- B. 2 and 3
- C. 3 and 4
- D. 2 and 4 Discuss
Correct Answer: 2 and 3
Explanation:
(2) - This is a Java keyword(3) - This is a Java keyword
(1) - Is incorrect because although it is a method of Thread/Runnable it is not a keyword
(4) - This is not a Java keyword the keyword is implements
- 5. What will be the output of the program?
class Equals { public static void main(String [] args) { int x = 100; double y = 100.1; boolean b = (x = y); /* Line 7 */ System.out.println(b); } }
Options- A. true
- B. false
- C. Compilation fails
- D. An exception is thrown at runtime Discuss
Correct Answer: Compilation fails
Explanation:
The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables.Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.
- 6. Which interface does java.util.Hashtable implement?
Options- A. Java.util.Map
- B. Java.util.List
- C. Java.util.HashTable
- D. Java.util.Collection Discuss
Correct Answer: Java.util.Map
Explanation:
Hash table based implementation of the Map interface.- 7. Which two of the following methods are defined in class Thread?
- start()
- wait()
- notify()
- run()
- terminate()
Options- A. 1 and 4
- B. 2 and 3
- C. 3 and 4
- D. 2 and 4 Discuss
Correct Answer: 1 and 4
Explanation:
(1) and (4). Only start() and run() are defined by the Thread class.(2) and (3) are incorrect because they are methods of the Object class. (5) is incorrect because there's no such method in any thread-related class.
- 8. What will be the output of the program?
class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); /* Line 5 */ t.run(); /* Line 6 */ } public void run() { for(int i=1; i < 3; ++i) { System.out.print(i + ".."); } } }
Options- A. This code will not compile due to line 5.
- B. This code will not compile due to line 6.
- C. 1..2..
- D. 1..2..3.. Discuss
Correct Answer: 1..2..
Explanation:
Line 6 calls the run() method, so the run() method executes as a normal method should and it prints "1..2.."A is incorrect because line 5 is the proper way to create an object.
B is incorrect because it is legal to call the run() method, even though this will not start a true thread of execution. The code after line 6 will not execute until the run() method is complete.
D is incorrect because the for loop only does two iterations.
- 9. 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 Discuss
Correct Answer: MyThread foo
Explanation:
Option 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"- 10. What will be the output of the program?
String s = "hello"; Object o = s; if( o.equals(s) ) { System.out.println("A"); } else { System.out.println("B"); } if( s.equals(o) ) { System.out.println("C"); } else { System.out.println("D"); }- A
- B
- C
- D
Options- A. 1 and 3
- B. 2 and 4
- C. 3 and 4
- D. 1 and 2 Discuss
Correct Answer: 1 and 3
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- 1. What will be the output of the program?