Re: [PATCH v2 1/4] rust: io: Add big-endian read and write functions

Danilo Krummrich
2026-02-08T17:52:23Z

On Sun Feb 8, 2026 at 6:17 PM CET, Daniel Almeida wrote:
>
>
>> On 5 Feb 2026, at 19:43, Danilo Krummrich <dakr@kernel.org> wrote:
>>
>> On Thu Feb 5, 2026 at 11:31 PM CET, Gary Guo wrote:
>>> On Thu Feb 5, 2026 at 7:05 PM GMT, Danilo Krummrich wrote:
>>>> (1) Devices are either little-endian or big-endian. Hence, having to write
>>>>
>>>> io.big_endian().write()
>>>>
>>>> is excessive, we always want big-endian for a big-endian device.
>>>
>>> You don't need to always write this. You just need to do `big_endian()` once
>>> when you obtain the io, and then keep using `BigEndian<Mmio>` instead of just
>>> `Mmio`, and the rest of code is still `.write()`.
>>
>> <snip>
>>
>>>> (2) It is error prone, if you forget to call big_endian() first, it is a bug.
>>>
>>> Moot point when `big_endian()` is only done once.
>>
>> Well, you need to do it at least once per driver entry point. For DRM IOCTLs for
>> instance you also have to consider that it is always Devres<Mmio>.
>>
>
> Well, this is also the case for relaxed(). I basically made peace with the fact that
>
> let mmio = mmio.relaxed();
> < use mmio >
>
> is going to be a reality per driver entrypoint, unless I misunderstood?

Well, there are two differences: Firstly, relaxed ordering should only be
possible in certain situations, but not always and not for every driver entry
point. Secondly, if you mistakenly forget it, you may suffer from a pretty
slight performance hit, but it is not going to be a bug.