How to get mutable references to two array elements at the same time?

by

You can solve this with split_at_mut():

let mut v = vec![1, 2, 3];
let (a, b) = v.split_at_mut(1);   // Returns (&mut [1], &mut [2, 3])
change(&mut a[0], &mut b[0]);

There are uncountably many safe things to do that the compiler unfortunately does not recognize yet. split_at_mut() is just like that, a safe abstraction implemented with an unsafe block internally.

We can do that too, for this problem. The following is something I use in code where I need to separate all three cases anyway (I: Index out of bounds, II: Indices equal, III: Separate indices).

enum Pair<T> {
    Both(T, T),
    One(T),
    None,
}

fn index_twice<T>(slc: &mut [T], a: usize, b: usize) -> Pair<&mut T> {
    if a == b {
        slc.get_mut(a).map_or(Pair::None, Pair::One)
    } else {
        if a >= slc.len() || b >= slc.len() {
            Pair::None
        } else {
            // safe because a, b are in bounds and distinct
            unsafe {
                let ar = &mut *(slc.get_unchecked_mut(a) as *mut _);
                let br = &mut *(slc.get_unchecked_mut(b) as *mut _);
                Pair::Both(ar, br)
            }
        }
    }
}

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