Multiple choice technology programming languages

class Fizz { int x = 5; public static void main(String[] args) { final Fizz f1 = new Fizz(); Fizz f2 = new Fizz(); Fizz f3 = FizzSwitch(f1,f2); System.out.println((f1 == f3) + " " + (f1.x == f3.x)); } static Fizz FizzSwitch(Fizz x, Fizz y) { final Fizz z = x; z.x = 6; return z; } } What is the result?

  1. true true

  2. false true

  3. true false

  4. false false

  5. Compilation fails.

  6. An exception is thrown at runtime.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The final keyword on f1 prevents reassigning the reference, but its fields can still be modified. z = x makes z refer to the same object as f1, so z.x = 6 changes f1.x to 6. Returning z returns the same object f1 refers to. Thus f1 == f3 is true (same reference) and f1.x == f3.x compares 6 == 6, also true.

AI explanation

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

Option A) true true - This option is correct because when comparing f1 and f3, the reference variables are pointing to the same object, f1, which was returned from the FizzSwitch method. Therefore, f1 == f3 is true. Additionally, the x variable in both f1 and f3 is modified to have a value of 6, so f1.x == f3.x is also true.

Option B) false true - This option is incorrect because f1 and f3 are pointing to the same object, so f1 == f3 should be true.

Option C) true false - This option is incorrect because f1.x and f3.x are both modified to have a value of 6, so f1.x == f3.x should be true.

Option D) false false - This option is incorrect because f1 and f3 are pointing to the same object, so f1 == f3 should be true. Additionally, f1.x and f3.x are both modified to have a value of 6, so f1.x == f3.x should also be true.

Option E) Compilation fails - This option is incorrect because there are no compilation errors in the given code.

Option F) An exception is thrown at runtime - This option is incorrect because there are no exceptions thrown in the given code.

The correct answer is A) true true. This option is correct because f1 and f3 are pointing to the same object, and the x variable in both objects is modified to have a value of 6.

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