Discussion
Home ‣ Java Programming ‣ Flow Control Comments
- Question
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
- Correct Answer
- 51
ExplanationIn 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.
Flow Control problems
Search Results
- 1. 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.- 2. What will be the output of the program?
public class Delta { static boolean foo(char c) { System.out.print(c); return true; } public static void main( String[] argv ) { int i = 0; for (foo('A'); foo('B') && (i < 2); foo('C')) { i++; foo('D'); } } }
Options- A. ABDCBDCB
- B. ABCDABCD
- C. Compilation fails.
- D. An exception is thrown at runtime. Discuss
Correct Answer: ABDCBDCB
Explanation:
'A' is only printed once at the very start as it is in the initialisation section of the for loop. The loop will only initialise that once.'B' is printed as it is part of the test carried out in order to run the loop.
'D' is printed as it is in the loop.
'C' is printed as it is in the increment section of the loop and will 'increment' only at the end of each loop. Here ends the first loop. Again 'B' is printed as part of the loop test.
'D' is printed as it is in the loop.
'C' is printed as it 'increments' at the end of each loop.
Again 'B' is printed as part of the loop test. At this point the test fails because the other part of the test (i < 2) is no longer true. i has been increased in value by 1 for each loop with the line: i++;
This results in a printout of ABDCBDCB
- 3. What will be the output of the program?
public class SwitchTest { public static void main(String[] args) { System.out.println("value =" + switchIt(4)); } public static int switchIt(int x) { int j = 1; switch (x) { case l: j++; case 2: j++; case 3: j++; case 4: j++; case 5: j++; default: j++; } return j + x; } }
Options- A. value = 2
- B. value = 4
- C. value = 6
- D. value = 8 Discuss
Correct Answer: value = 8
Explanation:
Because there are no break statements, once the desired result is found, the program continues though each of the remaining options.- 4. What will be the output of the program?
int x = 3; int y = 1; if (x = y) /* Line 3 */ { System.out.println("x =" + x); }
Options- A. x = 1
- B. x = 3
- C. Compilation fails.
- D. The code runs with no output. Discuss
Correct Answer: Compilation fails.
Explanation:
Line 3 uses an assignment as opposed to comparison. Because of this, the if statement receives an integer value instead of a boolean. And so the compilation fails.- 5. What will be the output of the program?
int x = l, y = 6; while (y--) { x++; } System.out.println("x = " + x +" y = " + y);
Options- A. x = 6 y = 0
- B. x = 7 y = 0
- C. x = 6 y = -1
- D. Compilation fails. Discuss
Correct Answer: Compilation fails.
Explanation:
Compilation fails because the while loop demands a boolean argument for it's looping condition, but in the code, it's given an int argument.while(true) { //insert code here }
- 6. 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.- 7. 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.- 8. 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.
- 9. 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 Discuss
Correct Answer: Prints: true,false,false
Explanation:
The 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.- 10. 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.
Comments
There are no comments.
Programming
Copyright ©CuriousTab. All rights reserved.
- 1. What will be the output of the program?