that references don't implement Deref

You can see all the types that implement Deref , and &T is in that list:

impl<'a, T> Deref for &'a T where T: ?Sized

The non-obvious thing is that there is syntactical sugar being applied when you use the * operator with something that implements Deref . Check out this small example:

use std::ops::Deref; fn main() { let s: String = "hello".into(); let _: () = Deref::deref(&s); let _: () = *s; }

error[E0308]: mismatched types --> src/main.rs:5:17 | 5 | let _: () = Deref::deref(&s); | ^^^^^^^^^^^^^^^^ expected (), found &str | = note: expected type `()` found type `&str` error[E0308]: mismatched types --> src/main.rs:6:17 | 6 | let _: () = *s; | ^^ expected (), found str | = note: expected type `()` found type `str`

The explicit call to deref returns a &str , but the operator * returns a str . It's more like you are calling *Deref::deref(&s) , ignoring the implied infinite recursion.

Xirdus is correct in saying

If deref returned a value, it would either be useless because it would always move out, or have semantics that drastically differ from every other function

Although "useless" is a bit strong; it would still be useful for types that implement Copy .

See also: