Aptitude Test On C-programming (Pointers And Strings) Set -a
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Are the expression *ptr++ and ++*ptr are same?
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True
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False
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This Aptitude test on C-Programming focuses on pointers and strings. It assesses understanding through questions on pointer operations, structure access, and string manipulations, essential for proficiency in C programming.
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- 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?
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'.'
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'&'
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'*'
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'->'
Correct Answer
A. '->'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.Rate this question:
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- 3.
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; }
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16, 2, 2
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20, 2, 2
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20, 4, 4
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10, 2, 4
Correct Answer
A. 20, 4, 4Explanation
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.Rate this question:
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- 4.
Trace the output. int main() { char str[] = "basic"; char *s = str; printf("%s ", s++ +3); printf("%s",s); return 0; }
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Ic asic
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C basic
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Asic sic
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C c
Correct Answer
A. 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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- 5.
What will be the output of the program ? int main() { char *str; str = "%s"; printf(str, "K\n"); return 0; }
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No Output
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%s
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K
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Error
Correct Answer
A. 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.
What will be the output of the program ? int main() { printf(5+"Good Morning\n"); return 0; }
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Good Morning
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Morning
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Good
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M
Correct Answer
A. 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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- 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; }
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10
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11
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5
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6
Correct Answer
A. 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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- 8.
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; }
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None of above
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Unequal
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Equal
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Error
Correct Answer
A. 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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- 9.
Which of the following function is used to find the first occurrence of a given string in another string?
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Strnset()
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Strstr()
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Strchr()
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Strrchr()
Correct Answer
A. Strstr()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.Rate this question:
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- 10.
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; }
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30
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27
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9
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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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- 11.
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; }
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JCK
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J65K
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JAK
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JACK
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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- 12.
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; }
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7, 9, 11, 13, 15
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2, 15, 6, 8, 10
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2, 4, 6, 8, 10
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3, 1, -1, -3, -5
Correct Answer
A. 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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- 13.
What will be the output of the program ? int main() { char p[] = "%d\n"; p[1] = 'c'; printf(p, 65); return 0; }
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A
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A
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C
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65
Correct Answer
A. 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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- 14.
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; }
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No Output
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0
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The string is empty
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The string is not empty
Correct Answer
A. 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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- 15.
What would be the equivalent pointer expression for referring the array element a[i][j][k][l]
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((((a+i)+j)+k)+l)
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*(*(*(*(a+i)+j)+k)+l)
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(((a+i)+j)+k+l)
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*((a+i)+j+k+l)
Correct Answer
A. *(*(*(*(a+i)+j)+k)+l)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.Rate this question:
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Current Version
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May 21, 2024Quiz Edited by
ProProfs Editorial Team -
Aug 25, 2011Quiz Created by
Mayurkothawade
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