When pattern-matching, you can specify that you'd like to get a mutable reference to the contained value by using ref mut
:
let mut score = Some(42);
if let Some(ref mut s) = score {
&mut s;
}
However, the inner value is not mutable:
error[E0596]: cannot borrow immutable local variable `s` as mutable
--> src/main.rs:4:14
|
4 | &mut s;
| ^
| |
| cannot reborrow mutably
| try removing `&mut` here
I tried to add in another mut
, but that was not valid:
if let Some(mut ref mut s) = score {
&mut s;
}
error: the order of `mut` and `ref` is incorrect
--> src/main.rs:3:17
|
3 | if let Some(mut ref mut s) = score {
| ^^^^^^^ help: try switching the order: `ref mut`
error: expected identifier, found keyword `mut`
--> src/main.rs:3:25
|
3 | if let Some(mut ref mut s) = score {
| ^^^ expected identifier, found keyword
error: expected one of `)`, `,`, or `@`, found `s`
--> src/main.rs:3:29
|
3 | if let Some(mut ref mut s) = score {
| ^ expected one of `)`, `,`, or `@` here
A mutable reference is a borrow to any type mut T , allowing mutation of T through that reference. The below code illustrates the example of a mutable variable and then mutating its value through a mutable reference ref_i . fn main() {
Mutable References First, we change s to be mut . Then we create a mutable reference with &mut s where we call the change function, and update the function signature to accept a mutable reference with some_string: &mut String . This makes it very clear that the change function will mutate the value it borrows.
RefCell allows borrowing immutable or mutable reference to the contained value. It tracks the borrows at runtime, via borrow and borrow_mut . pub fn borrow(&self) -> Ref<T> pub fn borrow_mut(&self) -> RefMut<T> The method borrow grants temporary access to the contained value via immutable reference.
Not a direct answer, but possible workarounds
if let Some(ref mut s) = score {
let mut s = s;
&mut s;
}
#[derive(Debug)]
struct X;
enum Foo<T> {
Bar(T),
_Baz,
}
fn main() {
let mut score = Foo::Bar(X);
if let Foo::Bar(ref mut s) = score {
//let x = s;
//println!("{:?}", **x); ! not possible
let x = &mut &mut *s; // &mut &mut X
println!("{:?}", **x);
}
}
Option
specificallyif let Some(ref mut s) = score.as_mut() {
s; //:&mut &mut i32
}
if let Some(mut s) = score.as_mut() {
&mut s;
}
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