Java programming — Objects and references with methods returning the same instance: class Two { byte x; } class PassO { public static void main(String [] args) { PassO p = new PassO(); p.start(); } void start() { Two t = new Two(); System.out.print(t.x + " "); Two t2 = fix(t); System.out.println(t.x + " " + t2.x); } Two fix(Two tt) { tt.x = 42; return tt; } }

Introduction / Context:
This problem examines Java's pass-by-value of references and default field initialization for primitive types inside objects.

Given Data / Assumptions:

• Field Two.x is a byte and defaults to 0 in a new object.
• Method fix mutates the same Two instance and returns that same reference.

Concept / Approach:
Java passes object references by value. Mutating the object via any reference affects the single underlying object instance. Primitives inside objects default to zero if not explicitly set.

Step-by-Step Solution:

Verification / Alternative check:
Insert an identity check like (t == t2) to confirm both references point to the same object.

Why Other Options Are Wrong:
They assume separate copies or null/default misunderstandings that do not apply here.

Common Pitfalls:
Confusing Java with pass-by-reference semantics; Java always passes by value, but the value of a reference lets you mutate the object.

Final Answer:
0 42 42