Question about calling methods on Rc<T>

help
1

When calling methods on type T in a Rc how does rust know to de-reference the Rc for me?

For example:

let x = Rc::new(String::new());
let y = x.len();

In this case rust knows that I am trying to get the length of the String inside the Rc. How does this work? Is this the implicit deref coercion at work?

2

This is due to the Deref trait. In short, if T1: Deref<Target = T2>, then any &T1 can be treated as &T2.

3

thank you for your clarification :slight_smile:

4

And in x.len() there is an implicit &x taking place thanks to the . operator auto-ref: since x: Rc<String> has no inherent .len() method, nor one from a trait, then Rust will try to call .len() on &x: &Rc<String>, and that's when a Deref coercion happens so that .len() ends up called on &*x: &String.

1 Like
5

Wow, that is something I had not totally grasped before.

The "." in x.len() is going to keep deferencing, if it can, until it finds a len() function.

Which means I can write:

fn main () {
    use ::std::rc::Rc;
    
    let x: Rc<Rc<Rc<Rc<Rc<String>>>>> = Rc::new(Rc::new(Rc::new(Rc::new(Rc::new("Hello, World!".into())))));
    let y = x.len();
    println!("x.len() = {:?}", y);
}

Awesome!

Is there an obfuscated Rust competition yet by the way?

1 Like
6

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