Pointer Arithmetic

beginner15 min readLesson 43 of 148

p + 1 doesn't move one byte — it moves one ELEMENT, guided by the pointer's type.

+1 means "one element further"

int a[5] = {5, 10, 15, 20, 25};
int *p = a;          // &a[0]

printf("%d\n", *p);        // 5
printf("%d\n", *(p + 1));  // 10 — one INT further (4 bytes), not one byte
printf("%d\n", *(p + 2));  // 15

The compiler scales arithmetic by sizeof(type): for int*, +1 is +4 bytes; for double*, +1 is +8. That is why pointer types matter.

Indexing IS pointer arithmetic

a[i]  is defined as  *(a + i)

a[3], 3[a], *(a+3) — all the same address. The bracket notation exists for readability, not because it is a different mechanism.

Walking with a pointer

int *p = a;
int *end = a + 5;          // one PAST the last element — legal to hold
while (p < end) {
    printf("%d ", *p);
    p++;                    // advance one element
}

You may compute and compare pointers one past the end; dereferencing there is undefined behavior.

Subtraction gives distance

int *p = &a[2];
printf("%ld\\n", (long)(p - a));   // 2 — element count, not bytes

What is NOT legal

arithmetic between unrelated arrays: p1 - p2 where they point into different arrays — undefined
dereferencing one-past-the-end, or any address outside the array
scaling with pointers to incomplete objects

Now practice

Arithmetic WorkbenchFunctions computed with pure pointer arithmetic — no subscript brackets on the array.4 challenges · · ~15 min String Functions by HandRebuild strlen, strcpy-style copy, and concatenation from first principles.3 challenges · · ~15 min