UGCNET-June2015-II-11

Question

What is the output of the following program: (Assume that the appropriate preprocessor directives are included and there is not 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 1

Ans is D

printf("%C", *(&S[3])); = D
        printf("%s", S+4); = EFGH
        printf("%u", S); = 1000

Answer 2

Ans is DEFGH1000 none of the option matches...

Answer 3

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.