Understanding the memory layout of Rc<T> pointers

help
1

Hello,

I would like to understand the memory layout while executing the following piece of code, taken from the standard library example (Cloning references)

use std::rc::Rc;

let vec_var = vec![1.0, 2.0, 3.0];
let foo = Rc::new(vec_var);
let a = Rc::clone(&foo);
let b = Rc::clone(&foo);

I have imagined the memory layout like the following drawing. Is that correct?
Thank you!

memory_layout_Rc_clone
memory_layout_Rc_clone1649×816 7.79 KB

2

Not correct. There is a single count for the Vec object. Move count and weak to the heap, stored just before the (pointer, len, cap) triple.

3

Note that you don't have to guess or make any assumptions. The source code of the Rust standard library is freely available right from the official documentation site, so you can just read it.

It shows:

struct RcBox<T: ?Sized> {
    strong: Cell<usize>,
    weak: Cell<usize>,
    value: T,
}

...

pub struct Rc<T: ?Sized> {
    ptr: NonNull<RcBox<T>>,
    phantom: PhantomData<RcBox<T>>,
}

I.e., Rc is really just a pointer, and the value lives together with the two reference counts.

4

So, the correct memory layout drawing should be the following:

memory_layout_Rc_clone2
memory_layout_Rc_clone21649×816 6.17 KB

5

Hello and thanks a lot for your help, you've really thrown light on it.

So, continuing my thoughts in the above discussion the following drawing should depict the memory layout of the following code take from the Book (Listing 15-18: A definition of List that uses Rc)

enum List {
    Cons(i32, Rc<List>),
    Nil,
}

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

fn main() {
    let a = Rc::new(Cons(5, Rc::new(Cons(10, Rc::new(Nil)))));
    let b = Cons(3, Rc::clone(&a));
    let c = Cons(4, Rc::clone(&a));
}

Listing 15-18
Listing 15-182741×1271 41.6 KB

6

The exact order of the ptr, len, cap fields of a vector could be different than what you used (the order is not guaranteed and may change from one compile to another). But otherwise yes.