C++ (old iostream.h): trace the ternary, references, and side effects—what does this program output and why? #include<iostream.h> int CuriousTabFunction(int m) { m *= m; return ((10) * (m /= m)); } int main() { int c = 9, *d = &c, e; int &z = e; e = CuriousTabFunction(c-- % 3 ? ++*d : (*d *= *d)); z = z + e / 10; cout << c << " " << e; return 0; }

Introduction / Context:
This problem checks understanding of the conditional (ternary) operator, pointer indirection, pre/post increment semantics, and references in C++. The code also includes a function that normalizes a nonzero number to 1 via m /= m.

Given Data / Assumptions:

• Old-style headers (<iostream.h>) but otherwise standard semantics.
• Initial c = 9, d points to c.
• Reference z bound to e.

Concept / Approach:
Evaluate the ternary condition carefully, track post-decrement and compound assignments, then propagate values through the function and the reference. Integer division and sequence points matter for the final value of e.

Step-by-Step Solution:
1) Condition: c-- % 3. The value used is 9 % 3 = 0, then c becomes 8. 2) False branch executes: (*d *= *d). Since d points to c (currently 8), c becomes 64. The argument to the function is 64. 3) In CuriousTabFunction: m *= m makes m = 4096; m /= m makes m = 1; returns 10 * 1 = 10. 4) Back in main: e = 10. Then, with z referencing e, z = z + e / 10 makes e = 10 + 1 = 11 (integer division). 5) Final print: c is 64; e is 11.

Verification / Alternative check:
Replace the function with a direct expression returning 10 to confirm that only the normalization to 1 affects the result, not the squared size.

Why Other Options Are Wrong:
Compile error: Code is valid in classic compilers.
64 9 / 64 10: Miss the final reference update by e/10.
8 11: Ignores that c was squared to 64.

Common Pitfalls:
Forgetting that c-- uses 9 in the expression; misunderstanding reference aliasing; overlooking integer division.

Final Answer:
64 11