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:
Note that all of the above is effectively true for Index
and IndexMut
as well.
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