In C programming (assume sizeof(int) = 2 bytes), what will this program print? #include void fun(char**); int main() { char argv[] = {"ab", "cd", "ef", "gh"}; fun(argv); return 0; } void fun(char p) { char t; t = (p += sizeof(int))[-1]; printf("%s ", t); } Carefully analyze the pointer arithmetic on the char parameter and identify the string printed.

Introduction / Context:This question tests pointer arithmetic with pointer-to-pointer types in C. The function receives a char that points to the first element of an array of char* (C-strings). Understanding how pointer increments scale by the size of the pointed-to type is critical for predicting the printed output.

Given Data / Assumptions:

• sizeof(int) = 2 bytes (as stated).
• Array: argv = { "ab", "cd", "ef", "gh" }.
• p is a char** initially pointing to &argv[0].
• Expression: t = (p += sizeof(int))[-1].

Concept / Approach:In C, pointer arithmetic advances by the size of the pointed-to type, not by bytes you write in code. Since p is a char**, p += 2 moves two elements forward (two pointers), not two bytes. Then the [-1] index selects the element immediately before the new position.

Step-by-Step Solution:Start: p → &argv[0].Compute p += sizeof(int) → p += 2 → p now points to &argv[2].Index [-1] → *(p - 1) → argv[1].Assign t = argv[1] → t points to "cd".printf prints "cd".

Verification / Alternative check:You can mentally rewrite (p += 2)[-1] as *(p + 2 - 1) with p originally at element 0, giving element 1 → "cd".

Why Other Options Are Wrong:"ab" is argv[0]; "ef" is argv[2]; "gh" is argv[3]; none match the indexed location after the pointer math. "Undefined behavior" does not apply because indexing stays within the array bounds.

Common Pitfalls:Confusing sizeof(int) with raw byte offset on a pointer-to-pointer; in C, pointer arithmetic already scales by element size. Also, misreading [-1] as invalid; it is valid when the pointer already advanced.

Final Answer:cd