Under what conditions can one pointer variable be subtracted from another?
Pointer subtraction isn't used very much, but can be handy to determine the distances between two array elements (i.e., the difference in the array indexes). You may not know exactly which element you're pointing to using pointer subtraction, but you can tell relative distances.
Pointer subtraction can subtract two pointers of the same type. The result is the distance (in array elements) between the two elements. Both pointers must point to objects that are members of the same array. After the compiler subtracts the addresses of two pointers, it divides the result (in bytes) by the size of the pointed-to object. Therefore, subtracting one pointer from another yields the number of elements between the two pointers. The formula used is rather simple. Assume that p1 and p2 are both pointers of type T *. Then, the value computed is:
( p2 - p1 ) == ( addr( p2 ) - addr( p1 ) ) / sizeof( T )
This can result in negative values if p2 has a smaller address than p1. p2 and p1 need not point to valid elements in an array. The formula above still works even when p2 and p1 contain invalid addresses (because they contain some address).
The following program illustrates how pointer subtraction yields the number of elements between two pointers. short s_array[10], *s_ptr1 = &s_array[0], *s_ptr2 = &s_array[3];
main()
{
printf("The address in s_ptr1 is %u\n", s_ptr1);
printf("The address in s_ptr2 is %u\n", s_ptr2);
printf("\nSubtracting s_ptr2 from s_ptr1 yields %d\n", s_ptr1 - s_ptr2);
printf("Subtracting s_ptr1 from s_ptr2 yields %d\n", s_ptr2 - s_ptr1);
}
The output from the preceding program was as follows: The address in s_ptr1 is 14737480
The address in s_ptr2 is 14737486
Subtracting s_ptr2 from s_ptr1 yields -3
Subtracting s_ptr1 from s_ptr2 yields 3
Notice that, when pointer s_ptr2 is subtracted from pointer s_ptr1, the pointer subtraction yields a negative value because s_ptr2 locates an object with a higher address than the object pointed to by s_ptr1.