GATE EXAM PREPARATION IN "C" - STRING

What does the following fragment of C-program print?
char c[] = "GATE2011"; 
char *p =c; 
printf("%s", p + p[3] - p[1]) ;

(A) GATE2011
(B) E2011
(C) 2011
(D) 011


Answer: (C) 

Explanation: See comments for explanation.

char c[] = "GATE2011"; 

 // p now has the base address string "GATE2011" 
char *p = c;  

// p[3] is 'E' and p[1] is 'A'. 
// p[3] - p[1] = ASCII value of 'E' - ASCII value of 'A' = 4 
// So the expression  p + p[3] - p[1] becomes p + 4 which is 
// base address of string "2011" 
printf("%s", p + p[3] - p[1]);  // prints 2011