Issue
I was reading the docs for Rust's Deref trait:
pub trait Deref {
type Target: ?Sized;
fn deref(&self) -> &Self::Target;
}
The type signature for the deref function seems counter-intuitive to me; why is the return type a reference? If references implement this trait so they can be dereferenced, what effect would this have at all?
The only explanation that I can come up with is that references don't implement Deref, but are considered "primitively dereferenceable". However, how would a polymorphic function which would work for any dereferenceable type, including both Deref<T> and &T, be written then?
Solution
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 (see docs).
If
derefreturned 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:
Note that all of the above is effectively true for Index and IndexMut as well.
Answered By - Shepmaster Answer Checked By - Mary Flores (PHPFixing Volunteer)
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