How to modify the last item of an array?

问题

Since arr is borrowed as mutable, the length of arr can't be gotten by calling len(). I'm stuck here, what's the right way to do it?

fn double_last(arr: &mut[i32]) -> &i32 {
    let last = &mut arr[arr.len() - 1];  // borrow checker error.
    //let last = &mut arr[3];            // fine
    *last *= 2;
    last
}

fn main() {
    let mut a = [1,2,3,4];
    println!("{}", double_last(&mut a));
    println!("{:?}", a);
}

回答1:

This will hopefully be fixed with the introduction of non-lexical lifetimes and the accompanying changes soon into the future (seems like it could be solved?).

For now though, you can satisfy the borrow checker by splitting that calculation out:

let n = arr.len() - 1;
let last = &mut arr[n];

回答2:

If you only need the last, you can use std::slice::last_mut

fn double_last(arr: &mut[i32]) -> &i32 {
    let last = arr.last_mut().unwrap();
    *last *= 2;
    last
}

来源：https://stackoverflow.com/questions/37043067/how-to-modify-the-last-item-of-an-array