GATE | GATE-CS-2015 (Set 1) | Question 45

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);
}

(A) 2036, 2036, 2036
(B) 2012, 4, 2204
(C) 2036, 10, 10
(D) 2012, 4, 6


Answer: (A)

Explanation:

x = 2000

Since x is considered as a pointer to an 
array of 3 integers and an integer takes 4
bytes, value of x + 3 = 2000 + 3*3*4 = 2036

The expression, *(x + 3) also prints same 
address as x is 2D array.


The expression *(x + 2) + 3 = 2000 + 2*3*4 + 3*4
                            = 2036

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