What would be the output of the following C program, assuming that: - The 2D array a has base address 1002 in memory - Each int occupies 2 bytes - The compiler prints addresses in decimal main() { int a[3][4] = { {1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12} }; printf("%u %u %u", a[0] + 1, *(a[0] + 1), *(*(a + 0) + 1)); }

Introduction / Context:
This question tests understanding of how multidimensional arrays are stored in memory in C and how pointer arithmetic works with such arrays. When you declare a 2D array of int values, the compiler lays them out contiguously in row major order. Expressions like a[0] + 1 and *(a[0] + 1) involve pointer arithmetic and dereferencing. The question also assumes specific starting addresses and a particular size of int to make the numeric output predictable.

Given Data / Assumptions:
- The array a is declared as int a[3][4] with initialiser values from 1 to 12 in row major form.
- The base address of a[0][0] is assumed to be 1002.
- Each int occupies exactly 2 bytes.
- The expression a[0] refers to the address of the first element in the first row, that is the address of a[0][0].
- The printf uses the %u format specifier to print addresses and integer values as unsigned integers.

Concept / Approach:
In C, a multidimensional array like int a[3][4] is stored as a contiguous block of 3 * 4 = 12 integers. The name a in expressions decays to a pointer to its first row, and a[0] is a pointer to the first element of that row. Pointer arithmetic on int pointers increments or decrements the address by the size of int in bytes. Dereferencing such expressions retrieves the stored integer values. To find the printed output, we compute the address and values step by step, using the given base address and element size.

Step-by-Step Solution:
1. The element a[0][0] has base address 1002. Because each int is 2 bytes, a pointer to the next element in the row will have address 1002 + 2 = 1004. 2. The expression a[0] is a pointer to a[0][0], so a[0] has the value 1002 when printed as an address. 3. The expression a[0] + 1 performs pointer arithmetic: it moves one int forward, adding 2 bytes. Thus a[0] + 1 has address 1004. 4. The expression *(a[0] + 1) dereferences the pointer a[0] + 1. That pointer refers to a[0][1], whose value from the initialiser list is 2. 5. The expression *(a + 0) is the same as a[0]; it is the pointer to the first element in the first row. 6. Therefore *( *(a + 0) + 1 ) is again equivalent to *(a[0] + 1), which is the value 2 stored at a[0][1]. 7. Putting it all together, printf prints the three values: address 1004, the value 2, and the value 2, separated by spaces.

Verification / Alternative check:
You can verify the address arithmetic by writing a simplified program that prints &a[0][0] and &a[0][1] using %p and comparing the difference. On a system where sizeof(int) is 2 bytes and the addresses start at 1002 for the sake of the example, the second address will be 1004. You can also confirm that a[0][1] always stores 2 by printing that element directly with printf("%d", a[0][1]);. These independent checks align with the pointer expressions used in the question and support the final result of 1004 2 2.

Why Other Options Are Wrong:
Option B, 1002 2 2, assumes that a[0] and a[0] + 1 point to the same address, which contradicts pointer arithmetic rules. Option C, 1004 3 3, incorrectly assumes that the value after the first element is 3 rather than 2, ignoring the actual initialiser. Option D, 1002 1 1, incorrectly uses the base address and the first element value for all three outputs. Only option A correctly reflects the address offset of 2 bytes and the repeated value 2 from the second element in the first row.

Common Pitfalls:
A common mistake is to think that incrementing a pointer by 1 adds one byte rather than one element size. Another frequent confusion is mixing up the role of a, a[0], and &a[0][0]. Learners sometimes also forget that in row major order, the first row is contiguous in memory, so a[0][0], a[0][1], a[0][2], and a[0][3] are consecutive elements. This question reinforces a careful step by step approach: identify the base element, use the correct element size, compute pointer arithmetic, and then dereference to obtain values.

Final Answer:
Given the base address 1002 and an int size of 2 bytes, the program prints 1004 2 2.