Pointer Arithmetic

Question 1

#include<stdio.h> int main() {     int arr[] = {10, 20, 30, 40, 50, 60};     int *ptr1 = arr;     int *ptr2 = arr + 5;     printf ("ptr2 - ptr1 = %d\n", ptr2 - ptr1);     printf ("(char*)ptr2 - (char*) ptr1 = %d",  (char*)ptr2 - (char*)ptr1);     getchar();     return 0; }

Output:

ptr2 - ptr1 = 5
  (char*)ptr2 - (char*) ptr1 = 20

In C, array name gives address of the first element in the array. So when we do ptr1 = arr, ptr1 starts pointing to address of first element of arr. Since array elements are accessed using pointer arithmetic, arr + 5 is a valid expression and gives the address of 6th element. Predicting value ptr2 – ptr1 is easy, it gives 5 as there are 5 inetegers between these two addresses. When we do (char *)ptr2, ptr2 is typecasted to char pointer. In expression “(int*)ptr2 – (int*)ptr1″, pointer arithmetic happens considering character poitners. Since size of a character is one byte, we get 5*sizeof(int) (which is 20) as difference of two pointers.

As an excercise, predict the output of following program.

#include<stdio.h> int main() {     char arr[] = "geeksforgeeks";     char *ptr1 = arr;     char *ptr2 = ptr1 + 3;     printf ("ptr2 - ptr1 = %d\n", ptr2 - ptr1);     printf ("(int*)ptr2 - (int*) ptr1 = %d",  (int*)ptr2 - (int*)ptr1);     getchar();     return 0; } Question 2: int fun(char *str1) {   char *str2 = str1;   while(*++str1);   return (str1-str2); }      int main() {   char *str = "geeksforgeeks";   printf("%d", fun(str));   getchar();   return 0; } Output: 13