In C, analyze multi-dimensional array initialization and deep dereferencing. What does this print? #include<stdio.h> int main() { int arr[3][3] = {1, 2, 3, 4}; printf("%d ", (((arr)))); return 0; }

Introduction / Context:
This problem tests understanding of how 2-D arrays are initialized and how dereferencing operators apply to array types and pointers. The expression uses three dereferences with parentheses that matter a lot.

Given Data / Assumptions:

• Declaration: int arr[3][3] = {1, 2, 3, 4}; which initializes arr[0] = {1,2,3}, arr[1] = {4,0,0}, arr[2] = {0,0,0}.
• Print statement: (((arr))).
• Standard C array-to-pointer decay rules apply in expressions.

Concept / Approach:
arr has type “array of 3 arrays of 3 int”. In expressions, arr usually decays to a pointer to its first element, i.e., int ()[3]. A single * applied to that pointer yields the first row, which is an array of 3 int. When used in expressions, that array further decays to int pointing to arr[0][0]. A second * then yields the int at arr[0][0] (which is 1). However, the third * attempts to dereference that int as if it were a pointer, which is illegal.

Step-by-Step Solution:
arr → decays to pointer to row: int ()[3].(arr) → the first row (type: array[3] of int), which in expression context decays to int*.((arr)) → the first int (arr[0][0]) as an lvalue int.((*(arr))) → applying * to an int → invalid indirection → compile-time error.

Verification / Alternative check:
Remove the last *: printf("%d", ((arr))); still incorrect type usage. Correct forms include printf("%d", **arr); which prints 1, because **arr equals arr[0][0].

Why Other Options Are Wrong:
They assume successful compilation and execution. Here the compiler will emit an invalid indirection error before any run occurs.

Common Pitfalls:
Forgetting that an int cannot be dereferenced; miscounting levels of indirection with array decay.

Final Answer:
Error: Invalid indirection