Difference of two addresses in C

Question

Possible Duplicate:
Pointer Arithmetic

The given code

int arr[12];
int * cur = arr;
cur++;
cout<<"cur-arr = "<<cur-arr<<endl;

Outputs 1, but I expected sizeof(int). Can someone please explain the nature of this behavior?

What don't you understand about that? What were you expecting? — Mat, Jan 04 '12 at 16:56
This is [pointer arithmetic](http://stackoverflow.com/questions/394767/pointer-arithmetic). — ulidtko, Jan 04 '12 at 16:57
All your questions that I've seen so far seem to be of the "confused, please explain" variety, but this is a Q&A site not a forum. So, ask questions. "I expected X because of Z, but I got Y. Why did I not get X instead?" — Lightness Races in Orbit, Jan 04 '12 at 16:59
@phonix: Almost certainly, but we want _him_ to tell us that. — Lightness Races in Orbit, Jan 04 '12 at 16:59
ya ... i had expected the sizeof (int) to be the output..anyways its clear now !!!! — Kunal, Jan 04 '12 at 17:06
@Kunal: So please take better care over your questions in the future. — Lightness Races in Orbit, Jan 04 '12 at 17:07
The better question is what made you think it returns `sizeof(int)`, since the book you learned from that you can subtract pointers pointing into the same array will surely have told you about basic pointer arithmetic... — PlasmaHH, Jan 04 '12 at 17:19

Alex · Answer 1 · 2012-01-05T00:06:02.173

4

It's a defined behavior of C pointer arithmetic. It uses the size of pointed type as a unit. If you change subtraction in the last line to

(char *)cur - (char *)arr

you get 4 in the output.

edited Jan 05 '12 at 00:06

answered Jan 04 '12 at 17:07

Alex

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That would be `char *`. You can't do arithmetic on `void *`. – Mike Seymour Jan 04 '12 at 17:23
Some compilers do allow `void *` arithmetic. `char *` is surely nicer. – ugoren Jan 04 '12 at 19:40

score 3 · Answer 2 · answered Jan 04 '12 at 16:58

This is the number of elements (ints here) between arr and cur (which is arr+1 at the time of subtraction). Compiler takes note that cur is a pointer to an integer and arr is an integer array. To get total number of bytes, try this:

(cur - arr) * sizeof(arr[0]);

score 2 · Answer 3 · answered Jan 04 '12 at 16:57

2

cur is a pointer to int, initialized to some value (arr - the semantics of array-to-pointer conversion are irrelevant here), incremented (cur++) and compared to its old value. Unsurprisingly, it grew by one through the increment operation.

Pointer arithmetic with a given type works just like regular arithmetic. While the pointer is advanced by sizeof(int) bytes in this example, the difference between pointers is also calculated in units of sizeof(int), so you see plain simple arithmetics.

answered Jan 04 '12 at 16:57

thiton

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That's not quite *incrementing*, the pointer itself grew not by one. It grew by `sizeof(int)`, but the pointer substraction operator returns difference in **elements**, not in bytes. – ulidtko Jan 04 '12 at 17:01
@ulidtko: It doesn't really matter. When you take the abstraction into account, it's certainly an increment, even though the underlying value increases by more than `(int)1`. – Lightness Races in Orbit Jan 04 '12 at 17:07

score 1 · Answer 4 · answered Jan 04 '12 at 17:07

1

Addition and substraction for pointers works in accordance to the pointer type.

answered Jan 04 '12 at 17:07

Alex

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Difference of two addresses in C

4 Answers4