Aptitude Test On C-programming (Pointers And Strings) Set -c
Questions and Answers
- 1.
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?
- A.
'.'
- B.
'&'
- C.
'*'
- D.
'->'
Correct Answer
D. '->'Explanation
The '->' operator is used to access data members of a structure through a pointer variable. This operator is used when we have a pointer to a structure and want to access its data members. It is used in the form of 'pointerVariable->member'.Rate this question:
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- 2.
What will be the output of the program ? int main() { inti=3, *j, k; j = &i; printf("%d\n", i**j*i+*j); return0; }
- A.
30
- B.
27
- C.
9
- D.
3
Correct Answer
A. 30Explanation
j=&i implies *j=i;
*j=3
i**j=3*3=9
i**j*i=9*3=27
i**j*i+*j=27+3=30Rate this question:
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- 3.
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; }
- A.
JACK
- B.
JCK
- C.
J65K
- D.
JAK
Correct Answer
A. JACKExplanation
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"Rate this question:
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- 4.
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; }
- A.
7, 9, 11, 13, 15
- B.
2, 4, 6, 8, 10
- C.
3, 1, -1, -3, -5
- D.
2, 15, 6, 8, 10
Correct Answer
D. 2, 15, 6, 8, 10Explanation
*(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+5Rate this question:
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- 5.
What will be the output of the program ? int main() { char *str; str = "%s"; printf(str, "K\n"); return 0; }
- A.
Error
- B.
K
- C.
No Output
- D.
%s
Correct Answer
B. KExplanation
printf(str, "K\n"); is replaced with printf("%s" , "K\n");
(since str = "%s";)
So it will print K.Rate this question:
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- 6.
Which of the following function is used to find the first occurrence of a given string in another string?
- A.
Strchr()
- B.
Strrchr()
- C.
Strstr()
- D.
Strnset()
Correct Answer
C. Strstr()Explanation
The function strstr() is used to find the first occurrence of a given string within another string. It searches for the substring in the main string and returns a pointer to the first occurrence of the substring. This function is commonly used in string manipulation and searching algorithms.Rate this question:
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- 7.
What will be the output of the program ? int main() { char p[] = "%d\n"; p[1] = 'c'; printf(p, 65); return 0; }
- A.
A
- B.
A
- C.
65
- D.
C
Correct Answer
B. AExplanation
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'.Rate this question:
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- 8.
Trace the output. int main() { char str[] = "basic"; char *s = str; printf("%s ", s++ +3); printf("%s",s); return 0; }
- A.
C c
- B.
Ic asic
- C.
Asic sic
- D.
C basic
Correct Answer
B. Ic asicExplanation
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''.Rate this question:
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- 9.
What would be the equivalent pointer expression for referring the array element a[i][j][k][l]
- A.
((((a+i)+j)+k)+l)
- B.
(((a+i)+j)+k+l)
- C.
*(*(*(*(a+i)+j)+k)+l)
- D.
*((a+i)+j+k+l)
Correct Answer
C. *(*(*(*(a+i)+j)+k)+l)Explanation
The correct answer is *(*(*(*(a+i)+j)+k)+l). This pointer expression is used to access the array element a[i][j][k][l]. The expression starts with a pointer to a, and then adds the offset i to access the correct element in the first dimension. It then adds the offset j to access the correct element in the second dimension. Similarly, it adds the offsets k and l to access the correct element in the third and fourth dimensions, respectively. Finally, the * operator is used to dereference the pointer and obtain the value of the array element.Rate this question:
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- 10.
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; }
- A.
The string is empty
- B.
The string is not empty
- C.
No output
- D.
0
Correct Answer
B. The string is not emptyExplanation
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".Rate this question:
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- 11.
Are the expression *ptr++ and ++*ptr are same?
- A.
True
- B.
False
Correct Answer
B. FalseExplanation
*ptr++ increments the pointer and not the value,
++*ptr increments the value being pointed by ptr .Rate this question:
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- 12.
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; }
- A.
10, 2, 4
- B.
20, 2, 2
- C.
20, 4, 4
- D.
16, 2, 2
Correct Answer
C. 20, 4, 4Explanation
The output of the program will be "20, 4, 4".
The sizeof operator is used to determine the size of a variable or data type. In this program, it is used to determine the size of the array "arr", the size of the pointer to the first element of the array "arr", and the size of each element in the array "arr".
Since the array "arr" has 5 elements and each element is of type int (which is 4 bytes in size), the size of the array is 5 * 4 = 20 bytes.
The sizeof(*arr) gives the size of the pointer to the first element of the array, which is 4 bytes in this case.
The sizeof(arr[0]) gives the size of each element in the array, which is also 4 bytes.Rate this question:
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- 13.
What will be the output of the program ? int main() { char str[] = "India\0\is Best\0"; printf("%d\n", strlen(str)); return 0; }
- A.
10
- B.
6
- C.
5
- D.
11
Correct Answer
C. 5Explanation
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"Rate this question:
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- 14.
What will be the output of the program ? int main() { printf(5+"Good Morning\n"); return 0; }
- A.
M
- B.
Morning
- C.
Good Morning
- D.
Good
Correct Answer
B. MorningExplanation
printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.
Hence the output is "Morning"Rate this question:
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- 15.
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; }
- A.
Equal
- B.
Error
- C.
Unequal
- D.
None of above
Correct Answer
C. UnequalExplanation
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".Rate this question:
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Mar 16, 2023Quiz Edited by
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Aug 25, 2011Quiz Created by
Mayurkothawade
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