What's the idiomatic way to copy from a primit

问题:

Consider the following snippet:

fn example(current_items: Vec<usize>, mut all_items: Vec<i32>) {
    for i in current_items.iter() {
        let mut result = all_items.get_mut(i);
    }
}

The compiler is complaining about i being &mut usize instead of usize:

error[E0277]: the trait bound `&usize: std::slice::SliceIndex<[()]>` is not satisfied
 --> src/lib.rs:3:36
  |
3 |         let mut result = all_items.get_mut(i);
  |                                    ^^^^^^^ slice indices are of type `usize` or ranges of `usize`
  |
  = help: the trait `std::slice::SliceIndex<[()]>` is not implemented for `&usize`

I've dug through the docs but the only way I see to satisfy the compiler is i.clone().

I'm definitely missing something obvious here. What's the idiomatic way to copy from primitive type reference by value?

回答1:

What's the idiomatic way to copy from primitive type reference by value?

You need to use * to dereference a reference.

let my_usize: usize = 5;
let ref_to_my_usize: &usize = &my_usize;
let copy_of_my_usize: usize = *ref_to_my_usize;

You can also dereference directly in your loop:

for &x in myvec.iter() {
//  ^-- note the &
}

回答2:

iter() on Vec<T> returns an iterator implementing Iterator<&T>, that is, this iterator will yield references into the vector. This is the most general behavior which allows convenient usage with non-copyable types.

However, primitive types (actually, any types which implement Copy trait) will be copied upon dereference anyway, so you just need this:

for i in current_items.iter() {
    let mut result = all_items.get_mut(*i);
}

Alternatively, you can use reference destructuring pattern:

for &i in current_items.iter() {
    let mut result = all_items.get_mut(i);
}

Now i is usize automatically and you don't need to dereference it manually.

回答3:

Don't use iter if you need an &mut reference. Use iter_mut instead.