Given: 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 true

  2. true false

  3. false true

  4. false false

  5. Compilation fails.

  6. An exception is thrown at runtime.

Show answer
Correct Option: A

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 it means that both the references f1 and f3 are pointing to the same object, and the value of x in that object is 6. So, (f1 == f3) is true because they refer to the same object and (f1.x == f3.x) is also true because the value of x is the same.

Option B) true false - This option is incorrect because (f1.x == f3.x) should be true since x is updated to 6 in the FizzSwitch method.

Option C) false true - This option is incorrect because (f1 == f3) should be true since f1 and f3 are referring to the same object.

Option D) false false - This option is incorrect because (f1 == f3) should be true since f1 and f3 are referring to the same object, and (f1.x == f3.x) should be true since x is updated to 6 in the FizzSwitch method.

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. This option is correct because (f1 == f3) is true because they refer to the same object and (f1.x == f3.x) is also true because the value of x is the same.

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