GATE CS 2011 Q22 — Programming in C

Let's analyze the given C program fragment step-by-step:

1. Array Initialization:

   char c[] = "GATE2011";

   This creates a character array c containing the string "GATE2011" followed by a null terminator '\0':
   • c[0] = 'G'
   • c[1] = 'A'
   • c[2] = 'T'
   • c[3] = 'E'
   • c[4] = '2'
   • c[5] = '0'
   • c[6] = '1'
   • c[7] = '1'
   • c[8] = '\0'
2. Pointer Assignment:

   char *p = c;

   The pointer p points to the first element of the array, i.e., &c[0].
3. Pointer Arithmetic: The expression in the printf statement is:

   $$\text{Expression} = p + p[3] - p[1]$$

   Substituting the values of p[3] and p[1]:
   • p[3] is 'E'
   • p[1] is 'A'
   Using their ASCII values (where 'E' = 69 and 'A' = 65):

   $$p[3] - p[1] = 'E' - 'A' = 69 - 65 = 4$$

   Therefore, the expression simplifies to:

   $$p + 4$$

   Since p points to c[0], p + 4 points to c[4].
4. Printing the String: The %s format specifier prints the string starting from the memory address p + 4 (which is &c[4]) up to the null terminator.

   The characters starting from index 4 are '2', '0', '1', '1', and '\0'.

   Thus, the output is 2011.