GATE CS 2018 Q.12 (CS Q.2): Pointers, Arrays and Strings

Consider the following C program.

#include<stdio.h> 
struct Ournode{ 
    char x,y,z; 
}; 
 
int main(){ 
    struct Ournode p = {'1', '0', 'a'+2}; 
    struct Ournode *q = &p; 
    printf ("%c, %c", *((char*)q+1), *((char*)q+2)); 
    return 0; 
}

The output of this program is:

0, c

0, a+2

'0', 'a+2'

'0', 'c'

Official answer and solution

Official answer: A
From: Official GATE 2018 answer key

Solution

Trace memory layout of the struct members and apply byte-level pointer arithmetic.

The structure struct Ournode contains three char members: x, y, and z. Since sizeof(char) == 1 and all members have 1-byte alignment, there is no padding between them:

Offset 0: x = '1'
Offset 1: y = '0'
Offset 2: z = 'a' + 2 = 'c' (since ASCII value of 'a' plus 2 is 'c')

Pointer q points to p. Cast (char*)q treats p as a pointer to the first byte:

((char*)q + 1) points to offset 1 (y), dereferencing gives '0'.
((char*)q + 2) points to offset 2 (z), dereferencing gives 'c'.

printf ("%c, %c", ...) prints the character values separated by comma and space, producing 0, c.

Final answer

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

Source

Paper: GATE 2018 Computer Science and Information Technology (CS)
Question: 12 of 65, printed as CS Q.2
Section: Computer Science and Information Technology

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