Why does second printf print 0

Question

#include<stdio.h>
int main()
{
    char arr[] = "somestring";
    char *ptr1 = arr;
    char *ptr2 = ptr1 + 3;
    printf("ptr2 - ptr1 = %ld\n", ptr2 - ptr1);
    printf("(int*)ptr2 - (int*) ptr1 = %ld",  (int*)ptr2 - (int*)ptr1);
    return 0;
}

I understand

 ptr2 - ptr1

gives 3 but cannot figure out why second printf prints 0.

@Jerry , I did not. This is from some C quiz I was taking. I got this wrong. — Ian McGrath, Oct 29 '13 at 11:58
@GrijeshChauhan, I am sorry. I did not understand what you are trying to say. — Ian McGrath, Oct 29 '13 at 12:00
@lan According to pointer mathematics when you add `1` then result will point to next location of same type Suppose `arr == 12` then `ptr1 == 12` and then `ptr1 + 3` == `15` and so `ptr2 == 15`. Now `ptr1` == `12` it you case to `int*` before adding `+` then `(int*)ptr1 + 1` = `16` for a system in which `sizeof (int*) == 4`. — Grijesh Chauhan, Oct 29 '13 at 12:03
Is this undefined behaviour (UB)? Dereferencing `(int*)ptr1` would be, as it's not aligned properly (depending on platform). But perhaps even the subtraction itself is UB with incorrectly-aligned pointers? — Aaron McDaid, Oct 29 '13 at 12:06
@AaronMcDaid, could be. But I tried on Mac,Windows and Linux. All gave the same answer. — Ian McGrath, Oct 29 '13 at 12:09

Benoit Blanchon · Accepted Answer · 2013-10-29T12:06:09.783

8

It's because when you substract two pointers, you get the distance between the pointer in number of elements, not in bytes.

(char*)ptr2-(char*)ptr1  // distance is 3*sizeof(char), ie 3
(int*)ptr2-(int*)ptr1  // distance is 0.75*sizeof(int), rounded to 0

EDIT: I was wrong by saying that the cast forces the pointer to be aligned

edited Oct 29 '13 at 12:06

answered Oct 29 '13 at 12:00

Benoit Blanchon

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Sizeof int* on my machine is 8bytes. yet ptr2=ptr1+5 gives 1. So I am assuming even if size is 8bytes, it can align on 4 byte boundary. Am I right? – Ian McGrath Oct 29 '13 at 12:03
@IanMcGrath `sizeof(int)` is 4, not `sizeof(int*)` – Yu Hao Oct 29 '13 at 12:04
@BenoitBlanchon, Now it makes perfect sense. Thanks. – Ian McGrath Oct 29 '13 at 12:08
Sorry about that, I went too fast. – Benoit Blanchon Oct 29 '13 at 12:09

score 4 · Answer 2 · edited May 23 '17 at 11:57

4

If you want to check the distance between addresses don't use (int *) or (void *), ptrdiff_t is a type able to represent the result of any valid pointer subtraction operation.

#include <stdio.h>
#include <stddef.h>

int main(void)
{
    char arr[] = "somestring";
    char *ptr1 = arr;
    char *ptr2 = ptr1 + 3;
    ptrdiff_t diff = ptr2 - ptr1;

    printf ("ptr2 - ptr1 = %td\n", diff);
    return 0;
}

EDIT: As pointed out by @chux, use "%td" character for ptrdiff_t.

edited May 23 '17 at 11:57

Community

1
1

answered Oct 29 '13 at 12:12

David Ranieri

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Did not know that. Have a +1, I accepted Benoit Blanchon's response because it explained me why it was 0. Still useful to know what you mentioned. – Ian McGrath Oct 29 '13 at 12:14
1

Per C11dr 6.5.6 9, the difference is type `ptrdiff_t`. Thus casting the difference to type `(ptrdiff_t)` does not change anything. To `printf()` this type, one should use `t` modifier as in `printf ("%td\n", ptr2 - ptr1)`. – chux - Reinstate Monica Oct 29 '13 at 14:42

score 1 · Answer 3 · answered Oct 29 '13 at 12:05

1

Casting a char pointer with int* would make it aligned to the 4bytes (considering int is 4 bytes here). Though ptr1 and ptr2 are 3 bytes away, casting them to int*, results in the same address -- hence the result.

answered Oct 29 '13 at 12:05

Siddhartha Ghosh

2,988
5
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score 0 · Answer 4 · edited Oct 29 '13 at 12:21

This is because sizeof(int) == 4

Each char takes 1 byte. Your array of chars looks like this in memory:

[s][o][m][e][s][t][r][i][n][g][0]

When you have an array of ints, each int occupies four bytes. storing '1' and '2' conceptually looks more like this:

[0][0][0][1][0][0][0][2]

Ints must therefore be aligned to 4-byte boundaries. Your compiler is aliasing the address to the lowest integer boundary. You'll note that if you use 4 instead of 3 this works as you expected.

The reason you have to perform a subtraction to get it to do it (just passing the casted pointers to printf doesn't do it) is because printf is not strictly typed, i.e. the %ld format does not contain the information that the parameter is an int pointer.

Why does second printf print 0

4 Answers4