In modern C++ (iostream), evaluate references bound to the same int and an enum-to-int assignment. What does this program print and why? #include<iostream> enum curioustab { a = 1, b, c }; int main() { int x = c; // x = 3 int &y = x; int &z = x; y = b; // x becomes 2 std::cout << z--; // print then decrement return 0; }

Introduction / Context: This problem mixes enum-to-int initialization, reference aliasing, and post-decrement printing behavior. All references y and z alias the same underlying int x.

Given Data / Assumptions:

• Enum values: a=1, b=2, c=3; x starts as 3.
• y and z both reference x.
• y=b sets x to 2 before the print.

Concept / Approach: With aliasing, assignments via any reference affect x. The expression z-- is post-decrement, so it prints the old value of x (which is 2) and only then decrements x to 1.

Step-by-Step Solution:

Verification / Alternative check: Replacing z-- with --z would print 1 immediately because pre-decrement decrements before yielding the value.

Why Other Options Are Wrong: 1 or 3 would require different timing of decrement or no prior assignment; compile error does not apply.

Common Pitfalls: Forgetting that all references alias the same int and misapplying pre/post decrement semantics.

Final Answer: The program will print the output 2.