What #[derive(IndexMut)] generates

Deriving IndexMut only works for a single field of a struct. Furthermore it requires that the type also implements Index, so usually Index should also be derived. The result is that you will mutably index it’s member directly.

With #[index_mut] or #[index_mut(ignore)] it’s possible to indicate the field that you want to derive IndexMut for.

1 Example usage

#[derive(Index, IndexMut)]
struct MyVec(Vec<i32>);

#[derive(Index, IndexMut)]
struct Numbers {
    #[index]
    #[index_mut]
    numbers: Vec<i32>,
    useless: bool,
}

fn main() {
    let mut myvec = MyVec(vec![5, 8]);
    myvec[0] = 50;
    assert_eq!(50, myvec[0]);

    let mut numbers = Numbers{numbers: vec![100, 200], useless: false};
    numbers[1] = 400;
    assert_eq!(400, numbers[1]);
}

2 Regular structs

When deriving IndexMut for a struct:

#[derive(Index, IndexMut)]
struct Numbers {
    #[index]
    #[index_mut]
    numbers: Vec<i32>,
    useless: bool,
}

Code like this will be generated to implement IndexMut:

impl<__IdxT> ::core::ops::IndexMut<__IdxT> for Numbers
where
    Vec<i32>: ::core::ops::IndexMut<__IdxT>,
{
    #[inline]
    fn index_mut(&mut self, idx: __IdxT) -> &mut Self::Output {
        <Vec<i32> as ::core::ops::IndexMut<__IdxT>>::index_mut(&mut self.numbers, idx)
    }
}

3 Enums

Deriving IndexMut is not supported for enums.