Rc<RefCell<T>>

Combine Rc and RefCell for mutable data with multiple owners.

Rc<RefCell<T>>

Rc shares but doesn't allow changes; RefCell changes but doesn't share. Combined, they give data that several owners can change:

use std::cell::RefCell;
use std::rc::Rc;

let ortak = Rc::new(RefCell::new(Vec::new()));
let a = Rc::clone(&ortak);
let b = Rc::clone(&ortak);
a.borrow_mut().push("a");
b.borrow_mut().push("b");
println!("{:?}", ortak.borrow());   // ["a", "b"]

It's common in structures like trees, graphs and observer lists.

The cycle danger

If two Rcs point at each other, their counters never reach zero and memory leaks. To prevent this, the link in the opposite direction is kept with Weak; Weak doesn't count as ownership:

let ebeveyn = Rc::new(RefCell::new(String::from("kök")));
let zayif = Rc::downgrade(&ebeveyn);
if let Some(e) = zayif.upgrade() {
    println!("{}", e.borrow());   // kök
}

upgrade() returns Some(Rc) if the value is still alive and None if it has been dropped.

Exercise Shared list

Create a shared list with Rc::new(RefCell::new(Vec::new())) and take two copies (a, b). Add "a" through a and "b" through b. Print the list and the number of owners separated by a space.

Show example solution

use std::cell::RefCell;
use std::rc::Rc;

fn main() {
    let ortak = Rc::new(RefCell::new(Vec::new()));
    let a = Rc::clone(&ortak);
    let b = Rc::clone(&ortak);
    a.borrow_mut().push("a");
    b.borrow_mut().push("b");
    println!("{:?} {}", ortak.borrow(), Rc::strong_count(&ortak));
}

Test yourself

Which combination does the same job across threads?
- Rc<RefCell<T>>
- Box<RefCell<T>>
- Arc<RefCell<T>>
- Arc<Mutex<T>>
Show answer

Arc<Mutex<T>> — RefCell isn't Sync; its thread-safe counterparts are Mutex and RwLock.

Shared and mutable

let ortak = Rc::new(____::new(Vec::new()));

Show answer

let ortak = Rc::new(RefCell::new(Vec::new()));

Type and capability

Rc<T>

Sharing
RefCell<T>

Interior mutation
Rc<RefCell<T>>

Shared mutation
Weak<T>

Non-owning link

Next lesson: Cell<T>

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