Re: [PATCH 0/3] Devres optimization with bound devices

From: Danilo Krummrich
Date: Sat Apr 26 2025 - 13:15:14 EST


On Sat, Apr 26, 2025 at 10:09:39AM -0700, Boqun Feng wrote:
> On Sat, Apr 26, 2025 at 03:30:38PM +0200, Danilo Krummrich wrote:
> > This patch series implements a direct accessor for the data stored within
> > a Devres container for cases where we can proof that we own a reference
> > to a Device<Bound> (i.e. a bound device) of the same device that was used
> > to create the corresponding Devres container.
> >
> > Usually, when accessing the data stored within a Devres container, it is
> > not clear whether the data has been revoked already due to the device
> > being unbound and, hence, we have to try whether the access is possible
> > and subsequently keep holding the RCU read lock for the duration of the
> > access.
> >
> > However, when we can proof that we hold a reference to Device<Bound>
> > matching the device the Devres container has been created with, we can
> > guarantee that the device is not unbound for the duration of the
> > lifetime of the Device<Bound> reference and, hence, it is not possible
> > for the data within the Devres container to be revoked.
> >
> > Therefore, in this case, we can bypass the atomic check and the RCU read
> > lock, which is a great optimization and simplification for drivers.
> >
>
> Nice! However, IIUC, if the users use Devres::new() to create a `Devres`
> , they will have a `Devres` they can revoke anytime, which means you can
> still revoke the `Devres` even if the device is bound.

No, a user of Devres can't revoke the inner Revocable itself. A user can only
drop the Devres instance, in which case the user also wouldn't be able to call
access_with() anymore.

> Also if a `Devres` belongs to device A, but someone passes device B's
> bound reference to `access_with()`, the compiler won't check for that,
> and the `Devres` can be being revoked as the same, no? If so the
> function is not safe.

Devres::access_with() compares the Device<Bound> parameter with its inner
ARef<Device>, and just fails if they don't match.