1. Given: 13. public class Pass { 14. public static void main(String [1 args) { 15. int x 5; 16. Pass p = new Pass(); 17. p.doStuff(x); 18. System.out.print(” main x = “+ x); 19. } 20. 21. void doStuff(int x) { 22. System.out.print(” doStuff x = “+ x++); 23. } 24. } What is the result?
technology programming languages

  1. Compilation fails.

  2. An exception is thrown at runtime

  3. doStuffx = 6 main x = 6

  4. doStuffx = 5 main x = 5

Show answer
Correct Option: D
Explanation:

To solve this question, the user needs to understand the basic concept of method invocation and parameter passing.

First, let's go through the code. The Pass class has a method main which initializes an integer variable x to 5. It then creates a new instance of the Pass class and invokes the doStuff method on that instance, passing in the value of x. Finally, it prints out the value of x.

The doStuff method takes an integer parameter x, prints out the value of x incremented by 1, and then increments the parameter x by 1.

Now, let's go through each option and explain why it is right or wrong:

A. Compilation fails - This option is incorrect. There are no syntax errors in the code.

B. An exception is thrown at runtime - This option is incorrect. There are no runtime errors in the code.

C. doStuffx = 6 main x = 6 - This option is incorrect. The value of x in main is still 5 after the doStuff method is called because Java is a pass-by-value language. The parameter x in doStuff is a copy of the value of x in main, so incrementing the parameter does not affect the original value of x.

D. doStuffx = 5 main x = 5 - This option is correct. The doStuff method prints out the value of x (which is 5) incremented by 1, resulting in the output "doStuff x = 6". However, the value of x in main remains 5 because Java is a pass-by-value language. Therefore, the final output is "main x = 5".

Therefore, the answer is: D. doStuffx = 5 main x = 5

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