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

[Boqun Feng]

On Mon, Sep 25, 2023 at 10:26:56PM +0000, Benno Lossin wrote:
> On 26.09.23 00:02, Boqun Feng wrote:
> > On Mon, Sep 25, 2023 at 11:58:46PM +0200, Alice Ryhl wrote:
> >> On 9/25/23 23:55, Boqun Feng wrote:
> >>> On Mon, Sep 25, 2023 at 09:03:52PM +0000, Benno Lossin wrote:
> >>>> On 25.09.23 20:51, Boqun Feng wrote:
> >>>>> On Mon, Sep 25, 2023 at 05:00:45PM +0000, Benno Lossin wrote:
> >>>>>> On 25.09.23 18:16, Boqun Feng wrote:
> >>>>>>> On Mon, Sep 25, 2023 at 03:07:44PM +0000, Benno Lossin wrote:
> >>>>>>>> ```rust
> >>>>>>>> struct MutatingDrop {
> >>>>>>>>          value: i32,
> >>>>>>>> }
> >>>>>>>>
> >>>>>>>> impl Drop for MutatingDrop {
> >>>>>>>>          fn drop(&mut self) {
> >>>>>>>>              self.value = 0;
> >>>>>>>>          }
> >>>>>>>> }
> >>>>>>>>
> >>>>>>>> let arc = Arc::new(MutatingDrop { value: 42 });
> >>>>>>>> let wr = arc.as_with_ref(); // this creates a shared `&` reference to the MutatingDrop
> >>>>>>>> let arc2: Arc<MutatingDrop> = wr.into(); // increments the reference count to 2
> >>>>>>>
> >>>>>>> More precisely, here we did a
> >>>>>>>
> >>>>>>> 	&WithRef<_> -> NonNull<WithRef<_>>
> >>>>>>>
> >>>>>>> conversion, and later on, we may use the `NonNull<WithRef<_>>` in
> >>>>>>> `drop` to get a `Box<WithRef<_>>`.
> >>>>>>
> >>>>>> Indeed.
> >>>>>>
> >>>>>
> >>>>> Can we workaround this issue by (ab)using the `UnsafeCell` inside
> >>>>> `WithRef<T>`?
> >>>>>
> >>>>> impl<T: ?Sized> From<&WithRef<T>> for Arc<T> {
> >>>>>        fn from(b: &WithRef<T>) -> Self {
> >>>>>            // SAFETY: The existence of the references proves that
> >>>>> 	// `b.refcount.get()` is a valid pointer to `WithRef<T>`.
> >>>>> 	let ptr = unsafe { NonNull::new_unchecked(b.refcount.get().cast::<WithRef<T>>()) };
> >>>>>
> >>>>> 	// SAFETY: see the SAFETY above `let ptr = ..` line.
> >>>>>            ManuallyDrop::new(unsafe { Arc::from_inner(ptr) })
> >>>>>                .deref()
> >>>>>                .clone()
> >>>>>        }
> >>>>> }
> >>>>>
> >>>>> This way, the raw pointer in the new Arc no longer derives from the
> >>>>> reference of `WithRef<T>`.
> >>>>
> >>>> No, the code above only obtains a pointer that has provenance valid
> >>>> for a `bindings::refcount_t` (or type with the same layout, such as
> >>>> `Opaque<bindings::refcount_t>`). But not the whole `WithRef<T>`, so accessing
> >>>> it by reading/writing will still be UB.
> >>>>
> >>>
> >>> Hmm... but we do the similar thing in `Arc::from_raw()`, right?
> >>>
> >>>       	pub unsafe fn from_raw(ptr: *const T) -> Self {
> >>> 	    ..
> >>> 	}
> >>>
> >>> , what we have is a pointer to T, and we construct a pointer to
> >>> `ArcInner<T>/WithRef<T>`, in that function. Because the `sub` on pointer
> >>> gets away from provenance? If so, we can also do a sub(0) in the above
> >>> code.
> >>
> >> Not sure what you mean. Operations on raw pointers leave provenance
> >> unchanged.
> > 
> > Let's look at the function from_raw(), the input is a pointer to T,
> > right? So you only have the provenance to T, but in that function, the
> > pointer is casted to a pointer to WithRef<T>/ArcInner<T>, that means you
> > have the provenance to the whole WithRef<T>/ArcInner<T>, right? My
> > question is: why isn't that a UB?
> 
> The pointer was originally derived by a call to `into_raw`:
> ```
>      pub fn into_raw(self) -> *const T {
>          let ptr = self.ptr.as_ptr();
>          core::mem::forget(self);
>          // SAFETY: The pointer is valid.
>          unsafe { core::ptr::addr_of!((*ptr).data) }
>      }
> ```
> So in this function the origin (also the origin of the provenance)
> of the pointer is `ptr` which is of type `NonNull<WithRef<T>>`.
> Raw pointers do not lose this provenance information when you cast
> it and when using `addr_of`/`addr_of_mut`. So provenance is something
> that is not really represented in the type system for raw pointers.

Ah, I see, that's the thing I was missing. Now it makes much sense to
me, thank you both!

> 
> When doing a round trip through a reference though, the provenance is
> newly assigned and thus would only be valid for a `T`:
> ```
> let raw = arc.into_raw();
> let reference = unsafe { &*raw };
> let raw: *const T = reference;
> let arc = unsafe { Arc::from_raw(raw) };
> ```

Agreed. This example demonstrates the key point: the provenances of raw
pointers are decided at derive time.

Regards,
Boqun


> Miri would complain about the above code.
> 
> -- 
> Cheers,
> Benno
> 
>