UGC-NET | UGC NET CS 2015 Jun – II | Question 11

What is the output of the following program?
(Assume that the appropriate preprocessor directives are included and there is no syntax error)

main ( ) 
     { char S[ ] = "ABCDEFGH"; 
      printf ("%C", *(&S[3]));
      printf ("%s", S + 4); 
      printf ("%u", S);        
      /* Base address of S is 1000 */        
     }

(A) ABCDEFGH1000
(B) CDEFGH1000
(C) DDEFGHH1000
(D) DEFGH1000


Answer: (D)

Explanation:

main ( ) 
     { char S[ ] = "ABCDEFGH"; 
      printf ("%C", *(&S[3]));
      printf ("%s", S + 4); 
      printf ("%u", S);        
      /* Base address of S is 1000 */        
     }

From above program:
printf (“%C”, *(&S[3])); will print character at *(&S[3]) i.e. D.
printf (“%s”, S + 4); will print string starting from S + 4 i.e. EFGH.
printf (“%u”, S); will print address of S i.e. 1000.
Since there is no new line instruction, So DEFGH1000 will be the output.
So, option (D) is correct.


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