In C, analyze the following program and determine the correct behavior (note the real line breaks):\n#include<stdio.h>\n#include<stdlib.h>\nint main()\n{\n int i = 0;\n i++;\n if(i <= 5)\n {\n printf("CuriousTab\n");\n exit(0);\n main();\n }\n return 0;\n}\n\nWhich statement best describes what actually happens when it runs?

Introduction / Context:
This item checks understanding of program termination using exit and control flow. It also explores whether a C program can recursively call main and what happens to statements after exit.

Given Data / Assumptions:

• i is incremented to 1 and satisfies i <= 5.
• printf prints one line when the if body runs.
• exit(0) is executed immediately after the print.
• A recursive call to main() appears after exit(0).

Concept / Approach:
exit terminates the entire process after running atexit handlers and flushing streams; no code placed after exit in the same control path executes. While calling main recursively is allowed by the standard, here that call is unreachable due to exit being unconditional.

Step-by-Step Solution:
Condition true → enter block;printf outputs “CuriousTab”.exit(0) terminates the process immediately.main() after exit is unreachable and never executes.

Verification / Alternative check:
Place a second printf after exit and observe it never runs; or substitute return to see different behavior.

Why Other Options Are Wrong:
Five prints would require looping or recursion; both are cut off by exit. “Once” is true but incomplete; option C captures the precise reasoning. There is no compile-time error solely for calling main; and “prints nothing” contradicts the first printf.

Common Pitfalls:
Confusing exit with return; assuming statements after exit can still run; believing main cannot be called recursively.

Final Answer:
The call to main() after exit() never executes