GO Classes DPP | C programming | Strings | UGC NET

Answer 1

Ans : D maybe😁

---

First Printf statement is Give Value at S[3] mean D

Second Give base address +4*(value of each string like for int ,float = 4 ) Give E to whole string EFGH

Third and last is base address of s mean 1000 and if you not solve 3rd statement also okay because in all option same value

 

and correct me if i am wrong

Answer 2

We can imagine it as given below, since it will be stored as a String only; S will be constant pointer to first element of the String;

A         B        C        D       E        F        G        H        \0

1000   1001   1002  1003  1004  1005   1006  1007   1008

(S+0)  (S+1)  (S+2) (S+3) (S+4) (S+5)   (S+6) (S+7)  (S+8)

Now we can say S[3] = *(S + 3) so we will evaluate it as       *(&*(1000 + 3))

                                                                                                *(&*(1003))

                                                                                                *(& D)  because *(1003) will be D

                                                                                                *(1003) which will be D

You can see the above string, S+4 is a pointer to E and since format specifier is %s the entire string EFGH will be printed so output till now is DEFGH

S is a constant pointer to first element of the string so it is 1000 only and we have to treat it as unsigned int so no problem at all

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.

Answer 4

S is a character array and since it is initialised using double quotes the final character is the null character \0. Character array S holds 9 characters in total.

• S is a constant pointer to the base address i.e 1000 (given)
• S+4 points to the address of the 5th element in character array S. Since we used %s as format specifier in printf , printf would print all the characters from this address until it encounters null character.
• *(&S[3]) dereferences the value at address of S[3] i.e 4th element of S. 

Using above information, answer should be option D.

 

Answer 5

 

---

 Given Code:

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

---

 STEP 1: Understand the array

char S[] = "ABCDEFGH";

Means:

Index:   0 1 2 3 4 5 6 7 8
Value:   A B C D E F G H \0   ← '\0' is automatically added
Address:1000 1001 1002 ... (assumed)

So:

• S[0] = 'A'

• S[3] = 'D'

• S[4] = 'E'

 STEP 2: First printf → printf("%C", *(&S[3]));

• S[3] = 'D'

• &S[3] = address of 'D'

• *(&S[3]) = value at address of 'D' = 'D'

 Output: D

 STEP 3: Second printf → printf("%s", S+4);

• S+4 = address pointing to S[4] = 'E'

• "%s" prints characters from this address until '\0'

So:

S+4 points to "EFGH"

Output: EFGH

 STEP 4: Third printf → printf("%u", S);

• S is base address = 1000 (given)

• %u prints the unsigned integer (address)

 Output: 1000

---

 FINAL OUTPUT

DEFGH1000

1. Know how pointers & arrays work. S+4 means skip first 4 characters.

2. Know *(&S[3]) just gives value of S[3].

3. %s prints till null. %u prints address

 Summary Output:

D E F G H 1 0 0 0

 

Answer 6

D