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

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Message-ID: <0b7fc71c-b3c7-4c29-92a9-587daa46ad59@ryhl.io>
Date:   Mon, 25 Sep 2023 17:30:05 +0200
From:   Alice Ryhl <alice@...l.io>
To:     Boqun Feng <boqun.feng@...il.com>,
        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 9/25/23 17:17, Boqun Feng wrote:
> 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?
> 

No, the raw pointer in the Arc was created from the immutable reference, 
so the raw pointer has the same restrictions as the immutable reference did.

Alice