GATE CS 2015 Set 1 Q35 — Arrays, Stacks and Queues

To find the output of the printf statement, let's analyze the memory layout of the 2D array x.

Given:

• Base address of x = $2000$
• Size of an integer = $4$ bytes
• Array declaration: unsigned int x[4][3] (4 rows, 3 columns)

The size of each row of the array is: $$\text{Row size} = 3 \times 4\text{ bytes} = 12\text{ bytes}$$

Step-by-Step Evaluation:

1. First Expression: x + 3

   x is a pointer to an array of 3 integers (type: unsigned int (*)[3]). Performing pointer arithmetic on x scales by the size of a row ($12$ bytes).

   $$\text{Value} = \text{Base Address} + 3 \times \text{Row size} = 2000 + 3 \times 12 = 2036$$
2. Second Expression: *(x + 3)

   Dereferencing x + 3 yields a pointer to the first element of the 4th row (type: unsigned int *). Numerically, the address of the row itself and the address of its first element are identical.

   $$\text{Value} = 2036$$
3. Third Expression: *(x + 2) + 3

   First, *(x + 2) points to the first element of the 3rd row (index 2):

   $$\text{Address of } *(x + 2) = 2000 + 2 \times 12 = 2024$$

   Since *(x + 2) is of type unsigned int *, adding $3$ to it scales by the size of an integer ($4$ bytes):

   $$\text{Value} = 2024 + 3 \times 4 = 2024 + 12 = 2036$$

Thus, the output printed is 2036, 2036, 2036.