Hello Everyone🙋🏻‍♂️, I have doubt in following C code. Could anyone help me? 🙏🏻 Please... [Code is in discription]

Question

#include <stdio.h>

int main()
{

int x =4;
int* ptr=&x;
int* k= ptr++;

int r= ptr-k;

printf("r is %d",r);
return 0;
}

Output: r is 1

And I am thinking that output should be 4.

Answer 1

Rohit Rathod you're right that the difference in bytes is the size of an int. Subtracting 2 pointers gives you the difference not in bytes but in chunks of that type's size.
So to get difference in bytes you would do: (ptr - k) * sizeof(int)
Since a char is 1 byte, if we cast ptr and k to pointers to chars you would get the difference in bytes:

printf("%d", (char*)ptr - (char*)k); // 4

https://stackoverflow.com/questions/9855482/pointer-address-difference

Answer 2

int r= ptr-k;
r is the difference between pointers ptr and k.

int* ptr=&x;
Initially ptr points to x. For example, let's say x resides at the memory address 1000 (int chunks, not bytes).

int *k= ptr++;
Post-increment is functionally equivalent to:
int* k = ptr; // k = 1000
ptr = ptr + 1; // ptr = 1000 + 1 = 1001

Therefore r is 1:
int r= ptr-k; // r = 1001 - 1000 = 1

The value of x does not matter because you never dereference any pointers.

Answer 3

jtrh thank you so much brother for your answer.

But my point of view is that when a pointer is incremented, it actually increments by the number equal to the size of the data type for which it is a pointer. For Example: If an integer pointer that stores address 1000 is incremented, then it will increment by 4(size of an int) and the new address it will points to 1004.
That's why I am thinking that answer should be 4.

Answer 4

jtrh thank you so much. Now I understood.