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?
Trait std::ops::DerefUsed for immutable dereferencing operations, like *v . In addition to being used for explicit dereferencing operations with the (unary) * operator in immutable contexts, Deref is also used implicitly by the compiler in many circumstances. This mechanism is called ' Deref coercion'.
The dereference operator is also known as the indirection operator. Simply put, the dereferencing operator allows us to get the value stored in the memory address of a pointer. In Rust, we use the Deref trait to customize the behaviour of the dereferencing operator.
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).
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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