C control flow and goto scope rules:\nPoint out the error in this program that jumps to a label inside another function.\n\n#include <stdio.h>\nint main()\n{\n void fun();\n int i = 1;\n while (i <= 5)\n {\n printf("%d\n", i);\n if (i > 2)\n goto here;\n i++;\n }\n return 0;\n}\nvoid fun()\n{\n here:\n printf("It works");\n}

Introduction / Context:
This problem validates understanding of goto scope rules. A goto can only jump to a label that is in the same function and within the same translation unit in scope. Crossing function boundaries with a goto is prohibited by the C standard.

Given Data / Assumptions:

• A label here: is defined inside fun().
• main() attempts goto here;.
• Loop and print statements otherwise are ordinary.

Concept / Approach:
Labels in C have function scope. They are not global symbols; they are only visible and valid within the function in which they are defined. Therefore, goto here; inside main() cannot legally reference here: defined in fun(). Compilers will emit an error such as “label ‘here’ used but not defined” in main, or “label exists in a different function.”

Step-by-Step Solution:
Identify the label definition site: in fun().Identify the jump site: in main().Apply rule: labels have function scope → cross-function jump is invalid.Fix by moving the label into main() or restructuring code using functions/returns.

Verification / Alternative check:
Compiling yields an error. Refactor to if (i > 2) { fun(); break; } to achieve a similar effect legally.

Why Other Options Are Wrong:
fun() cannot be accessed — access is fine; the problem is the cross-function goto. No error — incorrect.

Common Pitfalls:
Misunderstanding label scope; using goto for inter-function control transfer instead of calls/returns or flags.

Final Answer:
Error: goto cannot takeover control to other function