I expect the following code to compile and print Foo(6)
as the value of b
, owning the reference to a
is dropped after the match block.
It seems related to this compiler error:
error[E0502]: cannot borrow `a` as immutable because it is also borrowed as mutable
--> src/main.rs:26:22
|
13 | let b = get_foo(&mut a);
| - mutable borrow occurs here
...
26 | println!("{:?}", a);
| ^ immutable borrow occurs here
27 | }
| - mutable borrow ends here
Dropping the value of b
doesn't work either, because it is partially moved:
error[E0382]: use of partially moved value: `b`
--> src/main.rs:24:10
|
18 | Some(value) => *value = y,
| ----- value moved here
...
24 | drop(b);
| ^ value used here after move
|
= note: move occurs because `(b:std::prelude::v1::Some).0` has type `&mut u32`, which does not implement the `Copy` trait
Is there a better way to fix this rather than putting lines let b
and match b
into an inner block? That just looks wierd and ugly.
Shouldn't the compiler understand that the reference is dropped, and be able to compile that code?
#[derive(Debug)]
struct Foo(u32);
fn get_foo(bar: &mut Foo) -> Option<&mut u32> {
Some(&mut bar.0)
}
pub fn test() {
let mut x = 5;
let mut y = 6;
let mut a = Foo(x);
// {
let b = get_foo(&mut a);
match b {
Some(value) => *value = y,
_ => (),
}
// }
// drop(b);
println!("{:?}", a);
}
Yes, but not in stable Rust. You need non-lexical lifetimes:
Until then, just use the extra block.
drop
has nothing to do with borrows.See also: