Why can borrow_mut() be called on a Rc that contains a RefCell?

⚓ rust    📅 2025-06-05    👤 surdeus    👁️ 1      

surdeus

Hello,

The related questions I read do not answer this question.
I am reading the Rust book RefCell<T> and the Interior Mutability Pattern - The Rust Programming Language
and have a question on this code snippet:

#[derive(Debug)]
enum List {
    Cons(Rc<RefCell<i32>>, Rc<List>),
    Nil,
}

use crate::List::{Cons, Nil};
use std::cell::RefCell;
use std::rc::Rc;

fn main() {
    let value = Rc::new(RefCell::new(5));

    let a = Rc::new(Cons(Rc::clone(&value), Rc::new(Nil)));

    let b = Cons(Rc::new(RefCell::new(3)), Rc::clone(&a));
    let c = Cons(Rc::new(RefCell::new(4)), Rc::clone(&a));

    *value.borrow_mut() += 10;

    println!("a after = {a:?}");
    println!("b after = {b:?}");
    println!("c after = {c:?}");
}

To my understanding both Rc and RefCell are structs that among other things contain an element of generic type T. value is a Rc that contains a RefCell<i32> and borrow_mut is a method on a RefCell. So why is this method implemented for a Rc? I just checked this in the documentation. Is it because they want to make possible the use of boxing a RefCell inside it?

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