C | Pointer Basics | Question 6
#include<stdio.h> int main() { int arr[] = {10, 20, 30, 40, 50, 60}; int *ptr1 = arr; int *ptr2 = arr + 5; printf ( "Number of elements between two pointer are: %d." , (ptr2 - ptr1)); printf ( "Number of bytes between two pointers are: %d" , ( char *)ptr2 - ( char *) ptr1); return 0; } |
Assume that an int variable takes 4 bytes and a char variable takes 1 byte
(A) Number of elements between two pointer are: 5.
Number of bytes between two pointers are: 20
(B) Number of elements between two pointer are: 20.
Number of bytes between two pointers are: 20
(C) Number of elements between two pointer are: 5.
Number of bytes between two pointers are: 5
(D) Compiler Error
(E) Runtime Error
Answer: (A)
Explanation: Array name gives the address of first element in array. So when we do ‘*ptr1 = arr;’, ptr1 starts holding the address of element 10. ‘arr + 5’ gives the address of 6th element as arithmetic is done using pointers. So ‘ptr2-ptr1’ gives 5. When we do ‘(char *)ptr2’, ptr2 is type-casted to char pointer and size of character is one byte, pointer arithmetic happens considering character pointers. So we get 5*sizeof(int)/sizeof(char) as a difference of two pointers.
Please Login to comment...