I have a struct which contains both data and a writer which will eventually be used to write the data. The struct is wrapped in a RefCell
. Here\'s a small repro
You can manually invoke DerefMut
and then save the resulting reference:
fn write(s: RefCell<S>) {
let mut mut_s = s.borrow_mut();
let mut tmp = &mut *mut_s; // Here
let str = &tmp.data.string;
tmp.writer.write(str.as_bytes());
}
Or in one line:
fn write(s: RefCell<S>) {
let mut_s = &mut *s.borrow_mut(); // Here
let str = &mut_s.data.string;
mut_s.writer.write(str.as_bytes());
}
The problem is that borrow_mut doesn't return your struct directly — it returns a RefMut. Normally, this is transparent as this struct implements Deref and DerefMut, so any methods called on it are passed to the underlying type. The pseudo-expanded code looks something like this:
use std::cell::RefMut;
use std::ops::{Deref, DerefMut};
fn write(s: RefCell<S>) {
let mut mut_s: RefMut<S> = s.borrow_mut();
let str = &Deref::deref(&mut_s).data.string;
DerefMut::deref_mut(&mut mut_s).writer.write(str.as_bytes());
}
Rust doesn't track field-level borrows across function calls (even for Deref::deref
or DerefMut::deref_mut
). This causes your error, as the deref_mut
method would need to be called during the outstanding borrow from the previous Deref::deref
.
The expanded version with the explicit borrow looks like this, with a single call to Deref::deref_mut
:
use std::cell::RefMut;
use std::ops::DerefMut;
fn write(s: RefCell<S>) {
let mut mut_s: RefMut<S> = s.borrow_mut();
let tmp: &mut S = DerefMut::deref_mut(&mut mut_s);
let str = &tmp.data.string;
tmp.writer.write(str.as_bytes());
}
The compiler can then track that the two borrows from that temporary value are disjoint.
Note that this problem isn't unique to RefCell
! Any type that implements DerefMut can experience the same problem. Here's some from the standard library:
Box
MutexGuard
(from Mutex)PeekMut
(from BinaryHeap)RwLockWriteGuard
(from RwLock)String
Vec