C pointer-to-array allocation — total bytes allocated when int is 2 bytes.\n\n#include<stdio.h>\n#include<stdlib.h>\n#define MAXROW 3\n#define MAXCOL 4\n\nint main()\n{\n int (p)[MAXCOL];\n p = (int ()[MAXCOL]) malloc(MAXROW * sizeof(*p));\n return 0;\n}\n\nAssume sizeof(int) = 2. How many bytes does malloc request?

Introduction / Context:
This problem tests computing total bytes allocated when using a pointer to an array type with malloc. It reinforces that sizeof(*p) represents the entire array object, not just a single element.

Given Data / Assumptions:

• sizeof(int) = 2 bytes.
• *p has type int[4].
• We allocate MAXROW (3) blocks of type *p.

Concept / Approach:
sizeof(*p) = number_of_columns * sizeof(int) = 4 * 2 = 8 bytes. The malloc call requests MAXROW * sizeof(*p) = 3 * 8 = 24 bytes. The pointer type or casting does not change the size requested.

Step-by-Step Solution:

Verification / Alternative check:
Printing sizeof(*p) in code (as in the previous question) yields 8, confirming the per-row size. Multiplying by 3 confirms 24.

Why Other Options Are Wrong:

• 56/128: Overestimates unrelated to the given dimensions.
• 12: Would be 3 * 4, incorrectly using sizeof(int) instead of sizeof(*p).

Common Pitfalls:
Forgetting that *p is an entire array type and miscounting bytes by element instead of by row.

Final Answer:
24 bytes