Pointers - C Questions
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C Pointers Interview Questions
In most of the MNC interview questions such as in ZOHO interview question, IVTL Infoview interview questions, Amazon interview questions, GOOGLE interview questions, Infosys interview questions and even in Voonik interview questions, We come across several Tricky C Questions about which 2:5 of the questions are from pointers in c. Solving that kind of tricky C questions is not an easy task for all C programmers. We need more practices to solve it with ease. So we provide 25+ interesting C questions in pointers to make your MNC interview very easy.
Pointers Interview Questions
16. What will be the output of the C program?
#include<stdio.h>
int main(){
int *ptr;
*ptr = 5;
printf("%d" , *ptr);
return 0;
}
x
Option: B
Explanation
Pointer variable (*ptr) cannot be initialized.
17. What will be the output of the C program?
#include<stdio.h>
int main(){
int a = 36;
int *ptr;
ptr = &a;
printf("%u %u", *&ptr , &*ptr);
return 0;
}
x
Option: C
Explanation
& and * cancelled each other and display the address stored in a pointer variable ptr i.e) the address of a.
18. What will be the output of the C program?
#include<stdio.h>
int main(){
int num = 10;
printf("num = %d addresss of num = %u",num, &num);
num++;
printf("\n num = %d addresss of num = %u",num, &num);
return 0;
}
B. num = 10 address of num = 2293436
num = 11 address of num = 2293438 ✘
C. num = 10 address of num = 2293436
num = 11 address of num = 2293440 ✘
D. num = 10 address of num = 2293436
num = 11 address of num = 2293436 ✔
x
Option: D
Explanation
Here a variable num holds the value 10 and get its address as 2293436, then increment is done as num++ which uses the same address space to store the incremented value.
19. What will be the output of the C program?
#include<stdio.h>
int main(){
int i = 25;
int *j;
int **k;
j = &i;
k = &j;
printf("%u %u %u ",k,*k,**k);
return 0;
}
x
Option: A
Explanation
Here a pointer variable *j hold the address of i and then another pointer variable *k hold the address of j.
now
k = address of j
*k = address of i
**k = value of i.
20. What will be the output of the C program?
#include<stdio.h>
int main(){
int a, b, c;
char *p = 0;
int *q = 0;
double *r = 0;
a = (int)(p + 1);
b = (int)(q + 1);
c = (int)(r + 1);
printf("%d %d %d",a, b, c);
return 0;
}
x
Option: D
Explanation
Initializing pointer variable to zero is possible. Since initial address of any data type is zero, So its next address will be size of data type.
Here a pointer variable *p belongs to char data type, thus a = (int) (p + 1); increase its address by 1 as it belongs to char datatype. Thus variable a = 1 .
Here a pointer variable *q belongs to int data type, thus b = (int) (q + 1); increase its address by 4 as it belongs to int datatype. Thus variable b = 4 .
Here a pointer variable *r belongs to double data type, thus c = (int) (r + 1); increase its address by 8 as it belongs to double datatype. Thus variable c = 8 .
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