What is the output of the following C code? Assume that the address of $x$ is 2000 (in decimal) and an integer requires four bytes of memory.int main () { unsigned int x[4][3] = {{1,2,3},{4,5,6},{7,8,9},{10,11,12}}; printf("%u, %u, %u", x+3, *(x+3), *(x+2)+3); }

Given that x is a 2D array of type unsigned int[4][3] and the base address of x is $2000$. Each row contains 3 integers, taking $3 \times 4 = 12$ bytes.

1. Evaluating x + 3:

   • x has type unsigned int (*)[3] (pointer to an array of 3 integers).

   • x + 3 points to the 4th row (index 3).

   • Address $= 2000 + 3 \times \text{sizeof}(\text{unsigned int}[3]) = 2000 + 3 \times 12 = 2036$.

2. **Evaluating *(x + 3):**

   • *(x + 3) is the 4th row x[3], which decays to a pointer of type unsigned int * pointing to its first element x[3][0].

   • Its address value is the same as the start of the 4th row, which is $2036$.

3. **Evaluating *(x + 2) + 3:**

   • *(x + 2) is the 3rd row x[2], decaying to an unsigned int * pointing to x[2][0].

   • Base address of row 2 $= 2000 + 2 \times 12 = 2024$.

   • *(x + 2) + 3 performs pointer arithmetic on unsigned int *, which adds $3 \times \text{sizeof}(\text{unsigned int}) = 3 \times 4 = 12$ bytes.

   • Address $= 2024 + 12 = 2036$.

Therefore, printf("%u, %u, %u", ...) outputs 2036, 2036, 2036.