Aptitude test on C-Programming (Pointers and Strings) SET -A

1. Are the expression *ptr++ and ++*ptr are same?

True

False

*ptr++ increments the pointer and not the value,
++*ptr increments the value being pointed by ptr .

Explanation

*ptr++ increments the pointer and not the value,
++*ptr increments the value being pointed by ptr .

2. If the size of integer is 4bytes, What will be the output of the program?

#include

int main()
{
    int arr[] = {12, 13, 14, 15, 16};
    printf("%d, %d, %d\n", sizeof(arr), sizeof(*arr), sizeof(arr[0]));
    return 0;
}

16, 2, 2

20, 2, 2

20, 4, 4

10, 2, 4

The output of the program will be "20, 4, 4". The sizeof(arr) will give the total size of the array, which is 20 bytes since there are 5 integers in the array and each integer takes up 4 bytes of memory. The sizeof(*arr) will give the size of the first element in the array, which is 4 bytes. The sizeof(arr[0]) will give the size of the integer data type, which is also 4 bytes.

Explanation

The output of the program will be "20, 4, 4". The sizeof(arr) will give the total size of the array, which is 20 bytes since there are 5 integers in the array and each integer takes up 4 bytes of memory. The sizeof(*arr) will give the size of the first element in the array, which is 4 bytes. The sizeof(arr[0]) will give the size of the integer data type, which is also 4 bytes.

3. What will be the output of the program ? int main() { int i=3, *j, k; j = &i; printf("%d\n", i**j*i+*j); return 0; }

30

27

9

3

j=&i implies *j=i;
*j=3
i**j=3*3=9
i**j*i=9*3=27
i**j*i+*j=27+3=30

Explanation

j=&i implies *j=i;
*j=3
i**j=3*3=9
i**j*i=9*3=27
i**j*i+*j=27+3=30

4. What would be the equivalent pointer expression for referring the array element a[i][j][k][l]

((((a+i)+j)+k)+l)

*(*(*(*(a+i)+j)+k)+l)

(((a+i)+j)+k+l)

*((a+i)+j+k+l)

The correct answer is *(*(*(*(a+i)+j)+k)+l). This pointer expression is equivalent to referring to the array element a[i][j][k][l]. It uses nested dereferences and additions to access the desired element in the multi-dimensional array. The expression starts with (a+i) to access the ith element of the outermost array, then adds j to access the jth element of the next level array, and so on until the desired element is reached. Finally, the * operator is used to dereference the pointer and retrieve the value at that memory location.

Explanation

The correct answer is *(*(*(*(a+i)+j)+k)+l). This pointer expression is equivalent to referring to the array element a[i][j][k][l]. It uses nested dereferences and additions to access the desired element in the multi-dimensional array. The expression starts with (a+i) to access the ith element of the outermost array, then adds j to access the jth element of the next level array, and so on until the desired element is reached. Finally, the * operator is used to dereference the pointer and retrieve the value at that memory location.

5. If a variable is a pointer to a structure, then which of the following operator is used to access data members of the structure through the pointer variable?

'.'

'&'

'*'

'->'

The '->' operator is used to access data members of a structure through a pointer variable. This operator is used to dereference the pointer and access the data members of the structure it points to. It is commonly used when working with dynamic memory allocation or when passing structures as arguments to functions.

Explanation

The '->' operator is used to access data members of a structure through a pointer variable. This operator is used to dereference the pointer and access the data members of the structure it points to. It is commonly used when working with dynamic memory allocation or when passing structures as arguments to functions.

6. What will be the output of the program ?

int main()

{
    printf(5+"Good Morning\n");
    return 0;
}

Good Morning

Morning

Good

M

printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.
Hence the output is "Morning"

Explanation

printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.
Hence the output is "Morning"

7. What will be the output of the program ?

int main()

    char str[] = "India\0\is Best\0";

    printf("%d\n", strlen(str));

    return 0;

10

11

5

6

The function strlen returns the number of characters in the given string.
Therefore, strlen(str) becomes strlen("India") contains 5 characters. A string is a collection of characters terminated by '\0'.
The output of the program is "5"

Explanation

The function strlen returns the number of characters in the given string.
Therefore, strlen(str) becomes strlen("India") contains 5 characters. A string is a collection of characters terminated by '\0'.
The output of the program is "5"

8. What will be the output of the program ?



int main()
{
    char *str;
    str = "%s";
    printf(str, "K\n");
    return 0;
}

No Output

%s

K

Error

printf(str, "K\n"); is replaced with printf("%s" , "K\n");
(since str = "%s";)

So it will print K.

Explanation

printf(str, "K\n"); is replaced with printf("%s" , "K\n");
(since str = "%s";)

So it will print K.

9. Which of the following function is used to find the first occurrence of a given string in another string?

Strnset()

Strstr()

Strchr()

Strrchr()

The function strstr() is used to find the first occurrence of a given string within another string. It searches for a substring within a larger string and returns a pointer to the first occurrence of the substring. This function is commonly used in string manipulation and searching operations in programming.

Explanation

The function strstr() is used to find the first occurrence of a given string within another string. It searches for a substring within a larger string and returns a pointer to the first occurrence of the substring. This function is commonly used in string manipulation and searching operations in programming.

10. What will be the output of the program ? int main() { void *vp; char ch=74, *cp="JACK"; int j=65; vp=&ch; printf("%c", *(char*)vp); vp=&j; printf("%c", *(int*)vp); vp=cp; printf("%s", (char*)vp+2); return 0; }

JCK

J65K

JAK

JACK

Pointer always store integer value so cp will store the memory address of location where string "jack " is stored.

vp = &ch; Will store address of ch in vp so while we print content in printf it will print asccii value of 74 i.e "J"

vp = &j; It will assign address of j to vp again it will print ascii value of 65 as "A"

vp = cp; In this step cp is pointing to memory locatioon where string jack is stored and we r incrementing it by two so it will point to "C"
from sring "JACK" and since we hava given %S in printf so it will print content from c onward ie "CK"

Explanation

Pointer always store integer value so cp will store the memory address of location where string "jack " is stored.

vp = &ch; Will store address of ch in vp so while we print content in printf it will print asccii value of 74 i.e "J"

vp = &j; It will assign address of j to vp again it will print ascii value of 65 as "A"

vp = cp; In this step cp is pointing to memory locatioon where string jack is stored and we r incrementing it by two so it will point to "C"
from sring "JACK" and since we hava given %S in printf so it will print content from c onward ie "CK"

11. Trace the output. int main() { char str[] = "basic"; char *s = str; printf("%s ", s++ +3); printf("%s",s); return 0; }

Ic asic

C basic

Asic sic

C c

printf("%s ", s++ +3); -> ic since (s++ +3) and s++ is post increment but +3 just print string from 'i'. It ll not increment the pointer to 3.
printf("%s",s); here after incrementing (s++) It ll print ''asic''.

Explanation

printf("%s ", s++ +3); -> ic since (s++ +3) and s++ is post increment but +3 just print string from 'i'. It ll not increment the pointer to 3.
printf("%s",s); here after incrementing (s++) It ll print ''asic''.

12. What will be the output of the program ? int main() { char p[] = "%d\n"; p[1] = 'c'; printf(p, 65); return 0; }

A

C

65

char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d".
p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c".
printf(p, 65); becomes printf("%c", 65);
Therefore it prints the ASCII value of 65. The output is 'A'.

Explanation

char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d".
p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c".
printf(p, 65); becomes printf("%c", 65);
Therefore it prints the ASCII value of 65. The output is 'A'.

13. What will be the output of the program ? int main() { int i, a[] = {2, 4, 6, 8, 10}; change(a, 5); for(i=0; i<=4; i++) printf("%d, ", a[i]); return 0; } change(int *b, int n) { int i; for(i=0; i<n; i++) *(b+1) = *(b+i)+5; }

7, 9, 11, 13, 15

2, 15, 6, 8, 10

2, 4, 6, 8, 10

3, 1, -1, -3, -5

*(b+1) = *(b+i)+5;
1. i = 0 => *(b+1) ie 4 is replaced by *(b+0)+5 ie 2+5
2. i = 1 => *(b+1) ie 4 is replaced by *(b+1)+5 ie 4+5
3. i = 2 => *(b+1) ie 4 is replaced by *(b+2)+5 ie 6+5
4. i = 3 => *(b+1) ie 4 is replaced by *(b+3)+5 ie 8+5
5. i = 4 => *(b+1) ie 4 is replaced by *(b+4)+5 ie 10+5

Explanation

*(b+1) = *(b+i)+5;
1. i = 0 => *(b+1) ie 4 is replaced by *(b+0)+5 ie 2+5
2. i = 1 => *(b+1) ie 4 is replaced by *(b+1)+5 ie 4+5
3. i = 2 => *(b+1) ie 4 is replaced by *(b+2)+5 ie 6+5
4. i = 3 => *(b+1) ie 4 is replaced by *(b+3)+5 ie 8+5
5. i = 4 => *(b+1) ie 4 is replaced by *(b+4)+5 ie 10+5

14. What will be the output of the program ?

int main()

    char str1[] = "Hello";

    char str2[] = "Hello";

    if(str1 == str2)

        printf("Equal\n");

    else

        printf("Unequal\n");

    return 0;

None of above

Unequal

Equal

Error

if(str1 == str2) here the address of str1 and str2 are compared. The addresses of both variables are not same. Hence the if condition is failed.
At the else part it prints "Unequal".

Explanation

if(str1 == str2) here the address of str1 and str2 are compared. The addresses of both variables are not same. Hence the if condition is failed.
At the else part it prints "Unequal".

15. What will be the output of the program ? int main() { int i; char a[] = "\0"; if(printf("%s", a)) printf("The string is empty\n"); else printf("The string is not empty\n"); return 0; }

No Output

0

The string is empty

The string is not empty

The function printf() returns the number of characters printed on the console.
char a[] = "\0"; The variable a is declared as an array of characters and it initialized with "\0". It denotes that the string is empty.
if(printf("%s", a)) The printf() statement does not print anything, so it returns '0'(zero). Hence the if condition is failed.
In the else part it prints "The string is not empty".

Explanation

The function printf() returns the number of characters printed on the console.
char a[] = "\0"; The variable a is declared as an array of characters and it initialized with "\0". It denotes that the string is empty.
if(printf("%s", a)) The printf() statement does not print anything, so it returns '0'(zero). Hence the if condition is failed.
In the else part it prints "The string is not empty".

Aptitude Test On C-programming (Pointers And Strings) Set -a