Recursively calling main via a helper: trace argc from 1 up to 4 and print each value.\n\n#include <stdio.h>\nvoid fun(int);\n\nint main(int argc)\n{\n printf("%d ", argc);\n fun(argc);\n return 0;\n}\n\nvoid fun(int i)\n{\n if (i != 4)\n main(++i);\n}\n

Introduction / Context:
This question examines recursion through main and argument passing. Even though calling main recursively is unusual, it is permitted in C as a normal function call. The code prints the current argc each time before recursing until a stopping condition is reached.

Given Data / Assumptions:

• Assume normal program start with only the program name → initial argc = 1.
• fun(i) calls main(++i) until i == 4.
• Each call prints its argc with a trailing space.

Concept / Approach:
The execution is a recursion depth climb from 1 to 4. On each activation of main, the code prints its argc, then either recurses with argc + 1 or stops if argc == 4. No additional output occurs on unwinding.

Step-by-Step Solution:

Verification / Alternative check:
Replace fun with a loop printing i from 1 to 4 to confirm expected sequence; both methods output identical values.

Why Other Options Are Wrong:

• 1 2 3 / 2 3 4 / 1: Each omits at least one printed value or makes the wrong starting point.

Common Pitfalls:
Believing that main cannot be called like a normal function; forgetting the base condition i != 4 and the prefix increment semantics.

Final Answer:
1 2 3 4