GATE CS 2015 Set 1 Q.45 (CS Q.35): Pointers, Arrays and Strings

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.

#include <stdio.h>

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);
    return 0;
}

$2036, 2036, 2036$

$2012, 4, 2204$

$2036, 10, 10$

$2012, 4, 6$

Official answer and solution

Official answer: A
From: Official GATE 2015 answer key

Solution

Evaluate C 2D array expressions using pointer arithmetic scaling rules.

Given unsigned int x[4][3] with base address $2000$ , where each unsigned int occupies $4\text{ bytes}$ :

Each row contains $3$ unsigned integers, occupying $3 \times 4 = 12\text{ bytes}$ .
x has type unsigned int (*)[3] (pointer to an array of 3 ints).

x + 3:
Scales by the size of one row ( $12\text{ bytes}$ ):
$\text{Address} = 2000 + 3 \times 12 = 2000 + 36 = 2036.$
*(x + 3):
Dereferences x + 3 to yield the 1D array x[3], which decays to a pointer to its first element &x[3][0].
The numerical address of &x[3][0] is identical to the address of x[3]:
$\text{Address} = 2036.$
*(x + 2) + 3:
*(x + 2) is x[2], which points to &x[2][0] at address $2000 + 2 \times 12 = 2024$ .
Adding $+3$ advances by $3$ elements of type unsigned int ( $4\text{ bytes}$ each):
$\text{Address} = 2024 + 3 \times 4 = 2024 + 12 = 2036.$

All three formatted values printed by printf are 2036, 2036, 2036 (option A).

Final answer

Common trap

Assuming *(x + 3) dereferences the value of the integer rather than decaying to a pointer.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 1
Question: 45 of 65, printed as CS Q.35
Section: Computer Science and Information Technology

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