# FAQ: Pointers: Lesson - Pointer Arithmetic

This community-built FAQ covers the “Pointer Arithmetic” exercise from the lesson “Pointers: Lesson”.

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This exercise can be found in the following Codecademy content:

## FAQs on the exercise Pointer Arithmetic

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Hi fellow learners!
Unfortunately, I don’t see any value in pointer arithmetic so far.
Why would anyone want to change memory adresses in a program, so what’s the actual use of this?
And where would this be used?

1 Like

Hi _elwe!

I am also just learning this too, but from my understanding of C and arrays so far, this is really important for handling contiguous blocks of memory, like arrays. Here we have values stored consecutively in memory, e.g. the array `int myarray[] = {1, 2, 3, 4};` will use 4 bytes for each integer and store them consecutively in memory. This is really useful for quick access to different elements in the array when we use pointer arithmetic. If we want to get the element at the third index, we don’t need to move through the whole array to get there, we can just take the memory location of the first index, and use our pointer arithmetic as `pointer+= 3`, which here will move us 3*4=12 bytes on to reach element 3. This sounds trivial for a small array, but for really big arrays it’s a big time saver versus having to move through the whole array just to reach the last element, for instance (we would have to do this for some other data structures that don’t store elements in a contiguous block of memory, and so can’t take advantage of pointer arithmetic, like linked lists). So my (basic) understanding is that it’s useful for quick array access.

I hope that helps (and take it with a pinch of salt as I’m also learning, but this is my understanding for now )

For anyone that struggled with this too, I had written the code below, which ticked the boxes in the codeacademy exercise, but didn’t show any increment for the pointer when printing.

``````#include<stdio.h>

int main() {
double myvar = 12.1;
double* ptr1 = &myvar;
printf("%p\n", &ptr1);

// Code for Checkpoint 1 goes here:
// where I increment - removed so I don't give the whole solution
printf("%p\n", &ptr1);

// Code for Checkpoint 2 goes here:
// where I decrement - removed so I don't give the whole solution
printf("%p\n", &ptr1);
}
``````

Turns out I shouldn’t have used `&Ptr1` in my `printf` statement because this meant I was printing the memory address of the pointer itself, so the pointer of the pointer I suppose. So the correct print was just `printf("%p\n", ptr1);`