Takes slice ([ia..ib]) on a string slice (&str) returns a str

Question

I am learning Rust and I've encountered this when I was trying out examples:

fn main() {

    let x: &str = "hello";  // We created a string slice, `&str`
    let y: &str = x[2..4];  // I was expecting `y` to be `&str` as well
    // let y: str = x[2..4];  // `y` is actually a str (DST)!
    
    let z: &str = &x[2..4];  // This gives me what I expected!

    println!("str value: {}", z);
}

I am fairly confused in the first place; wasn't x a slice already? Why would I need to append the ampersand again in the last statement to make z a string slice (which in my mind it looks like &(&x) or just &&x)?

Related content that'd help you get grasp of the answer:

1. Why can't we move out of index (SO)
2. SO example of the impl of Index (SO)
3. Move and Copy Semantics in Rust (Rust Book)

Answer 1

I am a bit stunned in the first place; wasn't x a slice already?

Yes?

why would I need to append the ampersand again in the last statement to make z a string slice?

Because that's how indexing is defined in the language.

container[index] is actually syntactic sugar for *container.index(index)[...]. This allows nice things such as let value = v[index] if the type of value implements Copy.

And that's why you have to re-reference owned types or DSTs coming out of an indexing operation.