linux-kernel - Re: [PATCH v2 2/2] rust: arc: remove `ArcBorrow` in favour of `WithRef`

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Date:   Mon, 25 Sep 2023 08:17:48 -0700
From:   Boqun Feng <boqun.feng@...il.com>
To:     Alice Ryhl <aliceryhl@...gle.com>
Cc:     a.hindborg@...sung.com, alex.gaynor@...il.com,
        benno.lossin@...ton.me, bjorn3_gh@...tonmail.com, gary@...yguo.net,
        linux-kernel@...r.kernel.org, ojeda@...nel.org,
        rust-for-linux@...r.kernel.org, walmeida@...rosoft.com,
        wedsonaf@...il.com
Subject: Re: [PATCH v2 2/2] rust: arc: remove `ArcBorrow` in favour of
 `WithRef`

On Mon, Sep 25, 2023 at 03:00:47PM +0000, Alice Ryhl wrote:
> >>> I'm concerned about this change, because an `&WithRef<T>` only has
> >>> immutable permissions for the allocation. No pointer derived from it
> >>> may be used to modify the value in the Arc, however, the drop
> >>> implementation of Arc will do exactly that. 
> >> 
> >> That is indeed a problem. We could put the value in an `UnsafeCell`, but
> >> that would lose us niche optimizations and probably also other optimizations.
> >> 
> > 
> > Not sure I understand the problem here, why do we allow modifying the
> > value in the Arc if you only have a shared ownership?
> 
> Well, usually it's when you have exclusive access even though the value
> is in an `Arc`.
> 
> The main example of this is the destructor of the `Arc`. When the last
> refcount drops to zero, this gives you exclusive access. This lets you
> run the destructor. The destructor requires mutable access.
> 
> Another example would be converting the `Arc` back into an `UniqueArc`
> by checking that the refcount is 1. Once you have a `UniqueArc`, you can
> use it to mutate the inner value.
> 
> Finally, there are methods like `Arc::get_mut_unchecked`, where you
> unsafely assert that nobody else is using the value while you are
> modifying it. We don't have that in our version of `Arc` right now, but
> we might want to add it later.
> 

Hmm.. but the only way to get an `Arc` from `&WithRef` is

	impl From<&WithRef<T>> for Arc<T> {
	    ...
	}

, and we clone `Arc` in the that function (i.e. copying the raw
pointer), so we are still good?

Regards,
Boqun

> > Also I fail to see why `ArcBorrow` doesn't have the problem. Maybe I'm
> > missing something subtle here? Could you provide an example?
> 
> It's because `ArcBorrow` just has a raw pointer inside it. Immutable
> references give up write permissions, but raw pointers don't even if
> they are `*const T`.
> 
> Alice