C Programming Structure And Pointer! Trivia Questions Test

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 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 when the variable is a pointer to a structure, allowing us to access the members of the structure using the pointer. It is used to dereference the pointer and access the members of the structure directly.

Explanation

The '->' operator is used to access data members of a structure through a pointer variable. This operator is used when the variable is a pointer to a structure, allowing us to access the members of the structure using the pointer. It is used to dereference the pointer and access the members of the structure directly.

3. What will be the output of the program ? int main() { inti=3, *j, k; j = &i; printf("%d\n", i**j*i+*j); return0; }

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 expression is equivalent to referring to the array element a[i][j][k][l]. The expression starts with (a+i) which accesses the i-th element of the array a. Then, *(a+i) gives the pointer to the i-th element. Next, *(a+i)+j accesses the j-th element of the i-th element. This process continues with the addition of k and l, accessing the k-th and l-th elements respectively. Finally, the expression *(*(*(*(a+i)+j)+k)+l) dereferences the pointer to obtain the value of the array element a[i][j][k][l].

Explanation

The correct answer is *(*(*(*(a+i)+j)+k)+l). This expression is equivalent to referring to the array element a[i][j][k][l]. The expression starts with (a+i) which accesses the i-th element of the array a. Then, *(a+i) gives the pointer to the i-th element. Next, *(a+i)+j accesses the j-th element of the i-th element. This process continues with the addition of k and l, accessing the k-th and l-th elements respectively. Finally, the expression *(*(*(*(a+i)+j)+k)+l) dereferences the pointer to obtain the value of the array element a[i][j][k][l].

5. 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; }

JACK

JCK

J65K

JAK

The pointer always stores integer value so cp will store the memory address of the location where string "jack " is stored. VP = &ch; Will store address of ch in VP so while we print content in print it will print ASCII 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 location where string jack is stored and we r incrementing it by two so it will point to "C" from sring "JACK" and since we have given %S in print so it will print content from c onward ie "CK"

Explanation

The pointer always stores integer value so cp will store the memory address of the location where string "jack " is stored. VP = &ch; Will store address of ch in VP so while we print content in print it will print ASCII 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 location where string jack is stored and we r incrementing it by two so it will point to "C" from sring "JACK" and since we have given %S in print so it will print content from c onward ie "CK"

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

Strchr()

Strrchr()

Strstr()

Strnset()

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

Explanation

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

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


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

10, 2, 4

20, 2, 2

20, 4, 4

16, 2, 2

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 (5 integers, each taking 4 bytes). The sizeof(*arr) will give the size of the first element of the array, which is an integer and takes 4 bytes. The sizeof(arr[0]) will give the size of a single element in the array, which is also an integer and takes 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 (5 integers, each taking 4 bytes). The sizeof(*arr) will give the size of the first element of the array, which is an integer and takes 4 bytes. The sizeof(arr[0]) will give the size of a single element in the array, which is also an integer and takes 4 bytes.

8. What will be the output of the program ?

int main()

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

M

Morning

Good Morning

Good

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"

9. 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

6

5

11

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"

10. What will be the output of the program ?

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

Error

K

No Output

%s

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.

11. What will be the output of the program ? int main() { inti, 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, 4, 6, 8, 10

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

2, 15, 6, 8, 10

*(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

12. 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; }

The string is empty

The string is not empty

No output

0

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 the 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 the condition is failed. In the else part it prints "The string is not empty".

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

C c

Ic asic

Asic sic

C basic

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''.

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;

Equal

Error

Unequal

None of above

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

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

A

65

C

In this program, the character array p initially contains the format specifier "%d\n". Then, the character at index 1 of the array p is changed to 'c', so the array becomes "%c\n", indicating that it will print a character.

When printf(p, 65) is executed, it interprets the format specifier as "%c\n", so it expects a character argument. The integer value 65 is interpreted as the ASCII code for the character 'A'. Therefore, the output of the program will be:

A

Explanation

In this program, the character array p initially contains the format specifier "%d\n". Then, the character at index 1 of the array p is changed to 'c', so the array becomes "%c\n", indicating that it will print a character.

When printf(p, 65) is executed, it interprets the format specifier as "%c\n", so it expects a character argument. The integer value 65 is interpreted as the ASCII code for the character 'A'. Therefore, the output of the program will be:

A