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- Question
What will be the output of the program?
import java.util.*; class H { public static void main (String[] args) { Object x = new Vector().elements(); System.out.print((x instanceof Enumeration)+","); System.out.print((x instanceof Iterator)+","); System.out.print(x instanceof ListIterator); } }
Options- A. Prints: false,false,false
- B. Prints: false,false,true
- C. Prints: false,true,false
- D. Prints: true,false,false
- Correct Answer
- Prints: true,false,false
ExplanationThe Vector.elements method returns an Enumeration over the elements of the vector. Vector implements the List interface and extends AbstractList so it is also possible to get an Iterator over a Vector by invoking the iterator or listIterator method.
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- 1. What will be the output of the program?
public class Test { private static int[] x; public static void main(String[] args) { System.out.println(x[0]); } }
Options- A. 0
- B. null
- C. Compile Error
- D. NullPointerException at runtime Discuss
Correct Answer: NullPointerException at runtime
Explanation:
In the above code the array reference variable x has been declared but it has not been instantiated i.e. the new statement is missing, for example:private static int[]x = new int[5];
private static int[x] declares a static i.e. class level array.
the "new" keyword is the word that actually creates said array.
int[5] in association with the new sets the size of the array. so since the above code contains no new or size decalarations when you try and access x[0] you are trying to access a member of an array that has been declared but not intialized hence you get a NullPointerException at runtime.
- 2. What will be the output of the program?
TreeSet map = new TreeSet(); map.add("one"); map.add("two"); map.add("three"); map.add("four"); map.add("one"); Iterator it = map.iterator(); while (it.hasNext() ) { System.out.print( it.next() + " " ); }
Options- A. one two three four
- B. four three two one
- C. four one three two
- D. one two three four one Discuss
Correct Answer: four one three two
Explanation:
TreeSet assures no duplicate entries; also, when it is accessed it will return elements in natural order, which typically means alphabetical.- 3. 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 y: System.out.print("0 "); /* Line 11 */ case x-1: System.out.print("1 "); /* Line 12 */ case x: System.out.print("2 "); /* Line 13 */ } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. Compilation fails at line 11.
- D. Compilation fails at line 12. Discuss
Correct Answer: Compilation fails at line 11.
Explanation:
Case expressions must be constant expressions. Since x is marked final, lines 12 and 13 are legal; however y is not a final so the compiler will fail at line 11.- 4. What will be the output of the program?
public class If1 { static boolean b; public static void main(String [] args) { short hand = 42; if ( hand < 50 && !b ) /* Line 7 */ hand++; if ( hand > 50 ); /* Line 9 */ else if ( hand > 40 ) { hand += 7; hand++; } else --hand; System.out.println(hand); } }
Options- A. 41
- B. 42
- C. 50
- D. 51 Discuss
Correct Answer: 51
Explanation:
In Java, boolean instance variables are initialized to false, so the if test on line 7 is true and hand is incremented. Line 9 is legal syntax, a do nothing statement. The else-if is true so hand has 7 added to it and is then incremented.- 5. What will be the output of the program?
for (int i = 0; i < 4; i += 2) { System.out.print(i + " "); } System.out.println(i); /* Line 5 */
Options- A. 0 2 4
- B. 0 2 4 5
- C. 0 1 2 3 4
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails on the line 5 - System.out.println(i); as the variable i has only been declared within the for loop. It is not a recognised variable outside the code block of loop.- 6. What will be the output of the program?
package foo; import java.util.Vector; /* Line 2 */ private class MyVector extends Vector { int i = 1; /* Line 5 */ public MyVector() { i = 2; } } public class MyNewVector extends MyVector { public MyNewVector () { i = 4; /* Line 15 */ } public static void main (String args []) { MyVector v = new MyNewVector(); /* Line 19 */ } }
Options- A. Compilation will succeed.
- B. Compilation will fail at line 3.
- C. Compilation will fail at line 5.
- D. Compilation will fail at line 15. Discuss
Correct Answer: Compilation will fail at line 3.
Explanation:
Option B is correct. The compiler complains with the error "modifier private not allowed here". The class is created private and is being used by another class on line 19.- 7. What will be the output of the program?
public class Test { private static float[] f = new float[2]; public static void main (String[] args) { System.out.println("f[0] = " + f[0]); } }
Options- A. f[0] = 0
- B. f[0] = 0.0
- C. Compile Error
- D. Runtime Exception Discuss
Correct Answer: f[0] = 0.0
Explanation:
The choices are between Option A and B, what this question is really testing is your knowledge of default values of an initialized array. This is an array type float i.e. it is a type that uses decimal point numbers therefore its initial value will be 0.0 and not 0- 8. What will be the output of the program?
import java.util.*; class I { public static void main (String[] args) { Object i = new ArrayList().iterator(); System.out.print((i instanceof List)+","); System.out.print((i instanceof Iterator)+","); System.out.print(i instanceof ListIterator); } }
Options- A. Prints: false, false, false
- B. Prints: false, false, true
- C. Prints: false, true, false
- D. Prints: false, true, true Discuss
Correct Answer: Prints: false, true, false
Explanation:
The iterator() method returns an iterator over the elements in the list in proper sequence, it doesn't return a List or a ListIterator object.A ListIterator can be obtained by invoking the listIterator method.
- 9. What will be the output of the program?
public class Test { public static void main (String args[]) { String str = NULL; System.out.println(str); } }
Options- A. NULL
- B. Compile Error
- C. Code runs but no output
- D. Runtime Exception Discuss
Correct Answer: Compile Error
Explanation:
Option B is correct because to set the value of a String variable to null you must use "null" and not "NULL".- 10. What will be the output of the program?
public class Test { public static void main (String[] args) { String foo = args[1]; String bar = args[2]; String baz = args[3]; System.out.println("baz = " + baz); /* Line 8 */ } }And the command line invocation:> java Test red green blue
Options- A. baz =
- B. baz = null
- C. baz = blue
- D. Runtime Exception Discuss
Correct Answer: Runtime Exception
Explanation:
When running the program you entered 3 arguments "red", "green" and "blue". When dealing with arrays in java you must remember ALL ARRAYS IN JAVA ARE ZERO BASED therefore args[0] becomes "red", args[1] becomes "green" and args[2] becomes "blue".When the program entcounters line 8 above at runtime it looks for args[3] which has never been created therefore you get an
ArrayIndexOutOfBoundsException at runtime.
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- 1. What will be the output of the program?