Pointer subtraction confusion

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?

Solution 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!

Solution 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

Solution 3:

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

Solution 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.

Solution 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.