C++ reference to struct object: as both names alias the same , how do prints change before and after mutation? #include struct Tab { short n; }; int main() { Tab b; Tab &rb = b; b.n = 5; cout << b.n << " " << rb.n << " "; rb.n = 8; cout << b.n << " " << rb.n; return 0; }

Introduction / Context: This question reviews how references to objects work in C++. A reference is an alias: any update via the reference affects the same object and is visible through all aliases.

Given Data / Assumptions:

• Struct Tab has one member short n.
• b is an object; rb is a reference to b.

Concept / Approach: When you print b.n and rb.n before mutation, both show the same value. After assigning via rb.n, the shared underlying value changes and both names reflect the new value.

Step-by-Step Solution: 1) Set b.n = 5. 2) Print: b.n and rb.n both yield 5 (same object). 3) Assign rb.n = 8 (mutates the same underlying member). 4) Print again: both names now yield 8.

Verification / Alternative check: Change the second print to use only b.n; the value is still 8, demonstrating aliasing.

Why Other Options Are Wrong: 5 5 5 8 assumes distinct storage; 5 5 5 5 ignores the mutation; compile error is incorrect because code is valid.

Common Pitfalls: Assuming a reference is like a pointer requiring dereference with *; forgetting that a reference must be bound on initialization and cannot be reseated.

Final Answer: 5 5 8 8