Struct members double mutable borrow in Rust [duplicate] - rust

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 reproduction:
use std::cell::RefCell;
use std::io::Write;
struct Data {
string: String,
}
struct S {
data: Data,
writer: Vec<u8>,
}
fn write(s: RefCell<S>) {
let mut mut_s = s.borrow_mut();
let str = &mut_s.data.string;
mut_s.writer.write(str.as_bytes());
}
The compiler is angry:
error[E0502]: cannot borrow `mut_s` as mutable because it is also borrowed as immutable
--> src\main.rs:16:5
|
15 | let str = &mut_s.data.string;
| ----- immutable borrow occurs here
16 | mut_s.writer.write(str.as_bytes());
| ^^^^^ mutable borrow occurs here
17 | }
| - immutable borrow ends here
Is there a different API I should use?

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
Pin

Related

Deref and cannot borrow `form` as mutable more than once at a time [duplicate]

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 reproduction:
use std::cell::RefCell;
use std::io::Write;
struct Data {
string: String,
}
struct S {
data: Data,
writer: Vec<u8>,
}
fn write(s: RefCell<S>) {
let mut mut_s = s.borrow_mut();
let str = &mut_s.data.string;
mut_s.writer.write(str.as_bytes());
}
The compiler is angry:
error[E0502]: cannot borrow `mut_s` as mutable because it is also borrowed as immutable
--> src\main.rs:16:5
|
15 | let str = &mut_s.data.string;
| ----- immutable borrow occurs here
16 | mut_s.writer.write(str.as_bytes());
| ^^^^^ mutable borrow occurs here
17 | }
| - immutable borrow ends here
Is there a different API I should use?
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
Pin

Multiple borrow of disjoint fields of structure under mutex lock [duplicate]

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 reproduction:
use std::cell::RefCell;
use std::io::Write;
struct Data {
string: String,
}
struct S {
data: Data,
writer: Vec<u8>,
}
fn write(s: RefCell<S>) {
let mut mut_s = s.borrow_mut();
let str = &mut_s.data.string;
mut_s.writer.write(str.as_bytes());
}
The compiler is angry:
error[E0502]: cannot borrow `mut_s` as mutable because it is also borrowed as immutable
--> src\main.rs:16:5
|
15 | let str = &mut_s.data.string;
| ----- immutable borrow occurs here
16 | mut_s.writer.write(str.as_bytes());
| ^^^^^ mutable borrow occurs here
17 | }
| - immutable borrow ends here
Is there a different API I should use?
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
Pin

Why can I not call borrow_mut() after indexing into an immutable Vec<RefCell>?

Let's try to compile this code:
use std::cell::RefCell;
struct Foo {
v: Vec<RefCell<u8>>,
}
impl Foo {
fn f(&self, i: usize) {
let t = &mut *self.v[i].borrow_mut();
//let t = &mut *{self.v[i].borrow_mut()}; //compiled ok
}
}
fn main() {}
Compilation error:
error[E0596]: cannot borrow field `self.v` of immutable binding as mutable
--> src/main.rs:9:23
|
8 | fn f(&self, i: usize) {
| ----- use `&mut self` here to make mutable
9 | let t = &mut *self.v[i].borrow_mut();
| ^^^^^^ cannot mutably borrow field of immutable binding
Why does this code require adding &mut self to function signature in order to compile?
This is a known issue where IndexMut is sometimes selected when Index should actually be used.
Your workaround of using {} is reasonable, but you can also use Index explicitly:
use std::cell::RefCell;
fn f(v: Vec<RefCell<u8>>) {
use std::ops::Index;
let _t = &mut v.index(0).borrow_mut();
}
fn main() {}
See also:
Why does a mutable borrow of a closure through DerefMut not work?
How to use `BorrowMut` contained within `RefCell`?
Another workaround is to explicitly call RefCell::borrow_mut(&v[0]).

Error while trying to borrow 2 fields from a struct wrapped in RefCell

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 reproduction:
use std::cell::RefCell;
use std::io::Write;
struct Data {
string: String,
}
struct S {
data: Data,
writer: Vec<u8>,
}
fn write(s: RefCell<S>) {
let mut mut_s = s.borrow_mut();
let str = &mut_s.data.string;
mut_s.writer.write(str.as_bytes());
}
The compiler is angry:
error[E0502]: cannot borrow `mut_s` as mutable because it is also borrowed as immutable
--> src\main.rs:16:5
|
15 | let str = &mut_s.data.string;
| ----- immutable borrow occurs here
16 | mut_s.writer.write(str.as_bytes());
| ^^^^^ mutable borrow occurs here
17 | }
| - immutable borrow ends here
Is there a different API I should use?
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
Pin

Why is `&mut &foo` valid but `&mut a_ref_to_foo` is invalid?

In the following example, t1 compiles but t2 does not.
Is there anything special about &mut &stream? I don't think Deref kicks in.
use std::net::TcpStream;
fn t1() {
let stream = TcpStream::connect("127.0.0.1").unwrap();
let a = &mut &stream;
}
fn t2(stream: &TcpStream) {
let a = &mut stream;
}
Playground
9 | fn t2(stream: &TcpStream) {
| ------ use `mut stream` here to make mutable
10 | let a = &mut stream;
| ^^^^^^ cannot borrow mutably
&mut &foo
This does two things - it creates a temporary value of type &Foo, then creates another temporary of type &mut &Foo.
let a_ref_to_foo = &foo;
&mut a_ref_to_foo
This also creates a temporary of type &mut &Foo, but to something that is observable via the variable binding a_ref_to_foo, not another temporary.
The problem is the same as this example:
// Works
String::new().clear();
// Doesn't work
let s = String::new();
s.clear();
When you have a temporary value, you have ownership over it which includes treating it as mutable. Once it's been assigned to a binding, the binding has ownership and you must indicate that it's allowed to be mutated or not.

Resources