Tag: technology

Questions Related to technology

Given: 33. try { 34. // some code here 35. } catch (NullPointerException e1) { 36. System.out.print(”a”); 37. } catch (RuntimeException e2) { 38. System.out.print(”b”); 39. } finally { 40. System.out.print(”c”); 41. } What is the result if a NullPointerException occurs on line 34?

  1. c

  2. a

  3. ab

  4. ac

  5. bc

  6. abc


Correct Option: D

AI Explanation

To answer this question, let's go through each option to understand why it is correct or incorrect:

Option A) c - This option is incorrect because the "finally" block will always execute regardless of whether an exception is caught or not. Therefore, even if a NullPointerException occurs, "c" will be printed in addition to any other letters that may be printed.

Option B) a - This option is incorrect because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". However, the "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c".

Option C) ab - This option is incorrect because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". However, the "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c", but not "b".

Option D) ac - This option is correct because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". The "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c".

Option E) bc - This option is incorrect because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". However, the "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c", but not "b".

Option F) abc - This option is incorrect because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". The "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c". However, the catch block for RuntimeException (line 38) will not execute because the exception is specifically caught by the catch block for NullPointerException.

The correct answer is Option D) ac. This option is correct because if a NullPointerException occurs, the catch block for NullPointerException (line 36) will execute and print "a". The "finally" block will also execute and print "c". Therefore, the correct output will include both "a" and "c".

Given: 10. public class Fabric 11. public enum Color { 12. RED(0xff0000), GREEN(0x00ff00), BLUE(0x0000ff); 13. private final int rgb; 14. Color( int rgb) { this.rgb = rgb; } 15. public int getRGB() { return rgb; } 16. }; 17. public static void main( String[] argv) { 18. // insert code here 19. } 20. } Which two code fragments, inserted independently at line 18, allow the Fabric class to compile? (Choose two.)

  1. Color skyColor = BLUE;

  2. Color treeColor = Color.GREEN;

  3. Color purple = new Color( 0xff00ff);

  4. if( RED.getRGB() < BLUE.getRGB() ) {}

  5. Color purple = Color.BLUE + Color.RED;

  6. if( Color.RED.ordinal() < Color.BLUE.ordinal() ) {}


Correct Option: B,F

AI Explanation

To answer this question, let's review each of the given options:

Option A) Color skyColor = BLUE; This option assigns the enum constant "BLUE" to the variable "skyColor". This is incorrect because "BLUE" is a member of the "Color" enum and should be accessed as "Color.BLUE".

Option B) Color treeColor = Color.GREEN; This option correctly assigns the enum constant "GREEN" from the "Color" enum to the variable "treeColor". This is a valid way to access enum constants.

Option C) Color purple = new Color(0xff00ff); This option attempts to create a new enum constant using the constructor of the "Color" enum. However, enum constants are predefined and cannot be created using the constructor. This is incorrect.

Option D) if (RED.getRGB() < BLUE.getRGB()) {} This option correctly accesses the "getRGB()" method of the "RED" and "BLUE" enum constants and performs a comparison. This is a valid way to access enum constants and their methods.

Option E) Color purple = Color.BLUE + Color.RED; This option attempts to add the enum constants "BLUE" and "RED". However, enum constants cannot be added together using the "+" operator. This is incorrect.

Option F) if (Color.RED.ordinal() < Color.BLUE.ordinal()) {} This option correctly accesses the "ordinal()" method of the "RED" and "BLUE" enum constants and performs a comparison. This is a valid way to access enum constants and their methods.

Therefore, the correct options are B) Color treeColor = Color.GREEN; and F) if (Color.RED.ordinal() < Color.BLUE.ordinal()).

  1. Source file

  2. Directories

  3. Binary files

  4. All the above


Correct Option: D