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?

technology programming languages

  1. true false

  2. true true

  3. false true

  4. false false

Show answer
Correct Option: B

AI Explanation

To answer this question, let's go through each option:

Option A) true false - This option is incorrect. Let's analyze the code to understand why.

In the given code, we have two instances of the Fizz class, f1 and f2. The FizzSwitch method takes two Fizz objects as parameters, x and y. In the FizzSwitch method, the value of x is assigned to z, and then the value of x.x is changed to 6. Finally, z is returned.

In the main method, f1 is passed as the first parameter and f2 is passed as the second parameter to the FizzSwitch method. The returned value from the FizzSwitch method is assigned to f3.

Since z is assigned the value of x in the FizzSwitch method, and x is a reference to f1, z and f3 both reference the same object. Therefore, (f1 == f3) will evaluate to true.

Also, the value of x in f1 is changed to 6 in the FizzSwitch method. Therefore, (f1.x == f3.x) will also evaluate to true.

Option B) true true - This option is correct. As explained above, (f1 == f3) and (f1.x == f3.x) will both evaluate to true.

Option C) false true - This option is incorrect. As explained above, (f1 == f3) will evaluate to true.

Option D) false false - This option is incorrect. As explained above, (f1 == f3) will evaluate to true.

The correct answer is B) true true.

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