Arrays & pointers in C

Question

So I have the following code snippet:

#include <stdio.h>

void pointer_shift(int *a, int n);

int main(void) {
    int a[] = {100, 101, 102};
    pointer_shift(a1, 3);
}

void pointer_shift(int *a, int n) {
    int i;
    for (i = 0; i != n - 1; i++) {
        *(a + i) = *(a + i + 1);
    }
}

I just want to clarify how the pointers work in this snippet. So pointer_shift takes in 'a', a pointer to an int, correct? a1 is passed in to this parameter, and since arrays decay to a pointer to their first element, it works.

First of all, hopefully what I said in the above paragraph is correct. Secondly, what does *(a + i) = *(a + i + 1); actually do? Say we're on the first iteration of the for loop, and i = 0. Then the left side, *a, accesses what, exactly? Does it represent a pointer? I thought * was the dereferencing operator, and accesses the object that a pointer points to... And so then it sets *a = *(a + 1). (a + 1) is the next element in the array, but what exactly does this assignment do, and why?

Thanks!

Answer 1

It is actually not pointer shift, but value shift, *(a+i) is of same effect as a[i], so what it does is a[i] = a[i+1]