CODEWITHSUNDEEP
cpointers
Omar Khaled
Omar Khaled

Reputation: 25

The mechanism of pointers' subtraction

I have studied how math goes with pointers but there is a thing confusing me

for example :

int ages[]={10,20,30,40,50};
int* ages_ptr=ages;
for(ages_ptr;(ages_ptr-ages)<=4;ages_ptr++)
{
      printf("%d"*ages_ptr);
}
(ages_ptr-ages)<=4

here is my question how does result of ages_ptr-ages can be compared to int ?

Upvotes: 0

Views: 56

Answers (1)

rubikonx9
rubikonx9

Reputation: 1413

This line: int *ages_ptr = ages is roughly equivalent to: int *ages_ptr = &ages[0].

This means it it a pointer to the first element of the array. Actually, ages == &ages[0].

When you write ages_ptr++ it moves the pointer by sizeof(int) bytes, so to the next element of the array, so to &ages[1], &ages[2] and so on.

Pointers are just numbers - say, for example, value of &ages[0] is N. It has m elements, so the last element is at the adress of N + (m - 1) * sizeof(int).

Initially, value of *ages_ptr is N as well. On each iteration i it moves to N + i * sizeof(int).

In the loop condition you have: (ages_ptr - ages) <= 4. This 4 is actually m - 1, which makes it (ages_ptr - ages) <= m - 1. By substitution, we get: N + i - N <= m - 1, so i < m - 1, which is i < 4 in our case. (This 4 might be slightly misleading here, as coincidentally pretty often sizeof(int) == 4 as well.)

Notice how sizeof(int) was dropped in the substitution - because we are dealing with the pointers, the values of addition / subtraction are automatically handled for us. If you write a += x, you mean shift pointer a by x elements. If a is int *, it actually means a += 1 * sizeof(int) bytes.

The same goes for subtraction - that's why ages_ptr - ages returnes number of elements between the two pointers, not number of bytes.

Upvotes: 1

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