C++ (recursion with static indices and pre-recursion swapping) — determine the array order produced and printed. #include<iostream.h> struct CuriousTab { int arr[5]; public: void CuriousTabFunction(void); void Display(void); }; void CuriousTab::Display(void) { for (int i = 0; i < 5; i++) cout << arr[i] << " "; } void CuriousTab::CuriousTabFunction(void) { static int i = 0, j = 4; int tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; i++; j--; if (j != i) CuriousTabFunction(); } int main() { CuriousTab objCuriousTab = {{5, 6, 3, 9, 0}}; objCuriousTab.CuriousTabFunction(); objCuriousTab.Display(); return 0; } What output is produced?

Introduction / Context:
This follow-up uses the same static-index idea but performs the swap before recursion. The base condition is j != i, so recursion stops when the two indices meet at the center. You must trace the swaps carefully with persistent i, j to get the final array ordering.

Given Data / Assumptions:

• Initial array: [5, 6, 3, 9, 0].
• Static i = 0, j = 4.
• Each call swaps arr[i] with arr[j], then increments i and decrements j.

Concept / Approach:
Because the swap occurs immediately, we reverse pairs from the outermost inward, then stop when i == j. With static variables, each recursive frame sees updated indices rather than independent copies.

Step-by-Step Solution:

Verification / Alternative check:
Compare with the variant that swaps after recursion; you will notice the middle positions differ because of the different order of operations.

Why Other Options Are Wrong:

• Other sequences do not match the exact swap-before-recursion trace.

Common Pitfalls:
Thinking that recursion will reverse the entire array including the center twice; the stopping condition prevents that.

Final Answer:
0 9 3 6 5