Pointer subtraction confusion

Question

When we subtract a pointer from another pointer the difference is not equal to how many bytes they are apart but equal to how many integers (if pointing to integers) they are apart. Why so?

Answer 1

The idea is that you're pointing to blocks of memory

+----+----+----+----+----+----+
| 06 | 07 | 08 | 09 | 10 | 11 | mem
+----+----+----+----+----+----+
| 18 | 24 | 17 | 53 | -7 | 14 | data
+----+----+----+----+----+----+

If you have int* p = &(array[5]) then *p will be 14. Going p=p-3 would make *p be 17.

So if you have int* p = &(array[5]) and int *q = &(array[3]), then p-q should be 2, because the pointers are point to memory that are 2 blocks apart.

When dealing with raw memory (arrays, lists, maps, etc) draw lots of boxes! It really helps!

Answer 2

Because everything in pointer-land is about offsets. When you say:

int array[10];
array[7] = 42;

What you're actually saying in the second line is:

*( &array[0] + 7 ) = 42;

Literally translated as:

* = "what's at"
(
  & = "the address of"
  array[0] = "the first slot in array"
  plus 7
)
set that thing to 42

And if we can add 7 to make the offset point to the right place, we need to be able to have the opposite in place, otherwise we don't have symmetry in our math. If:

&array[0] + 7 == &array[7]

Then, for sanity and symmetry:

&array[7] - &array[0] == 7

Answer 3

So that the answer is the same even on platforms where integers are different lengths.

Answer 4

Say you have an array of 10 integers:

int intArray[10] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };

Then you take a pointer to intArray:

int *p = intArray;

Then you increment p:

p++;

What you would expect, because p starts at intArray[0], is for the incremented value of p to be intArray[1]. That's why pointer arithmetic works like that. See the code here.

Answer 5

"When you subtract two pointers, as long as they point into the same array, the result is the number of elements separating them"

Check for more here.

Answer 6

This way pointer subtraction behaves is consistent with the behaviour of pointer addition. It means that p1 + (p2 - p1) == p2 (where p1 and p2 are pointers into the same array).

Pointer addition (adding an integer to a pointer) behaves in a similar way: p1 + 1 gives you the address of the next item in the array, rather than the next byte in the array - which would be a fairly useless and unsafe thing to do.

The language could have been designed so that pointers are added and subtracted the same way as integers, but it would have meant writing pointer arithmetic differently, and having to take into account the size of the type pointed to:

• p2 = p1 + n * sizeof(*p1) instead of p2 = p1 + n
• n = (p2 - p1) / sizeof(*p1) instead of n = p2 - p1

So the result would be code that is longer, and harder to read, and easier to make mistakes in.

Answer 7

When applying arithmetic operations on pointers of a specific type, you always want the resulting pointer to point to a "valid" (meaning the right step size) memory-address relative to the original starting-point. That is a very comfortable way of accessing data in memory independently from the underlying architecture.

If you want to use a different "step-size" you can always cast the pointer to the desired type:

int a = 5;
int* pointer_int = &a;
double* pointer_double = (double*)pointer_int; /* totally useless in that case, but it works */

Answer 8

@fahad Pointer arithmetic goes by the size of the datatype it points.So when ur pointer is of type int you should expect pointer arithmetic in the size of int(4 bytes).Likewise for a char pointer all operations on the pointer will be in terms of 1 byte.