Re: [PATCH v2 2/2] io: prevent compiler reordering on the default readX() implementation

Arnd Bergmann
2018-04-03T11:13:31Z

On Tue, Apr 3, 2018 at 12:49 PM, Mark Rutland <mark.rutland@arm.com> wrote:
> Hi,
>
> On Fri, Mar 30, 2018 at 11:58:13AM -0400, Sinan Kaya wrote:
>> The default implementation of mapping readX() to __raw_readX() is wrong.
>> readX() has stronger ordering semantics. Compiler is allowed to reorder
>> __raw_readX().
>
> Could you please specify what the compiler is potentially reordering
> __raw_readX() against, and why this would be wrong?
>
> e.g. do we care about prior normal memory accesses, subsequent normal
> memory accesses, and/or other IO accesses?
>
> I assume that the asm-generic __raw_{read,write}X() implementations are
> all ordered w.r.t. each other (at least for a specific device).

I think that is correct: the compiler won't reorder those because of the
'volatile' pointer dereference, but it can reorder access to a normal
pointer against a __raw_readl()/__raw_writel(), which breaks the scenario
of using writel to trigger a DMA, or using a readl to see if a DMA has
completed.

The question is whether we should use a stronger barrier such
as rmb() amd wmb() here rather than a simple compiler barrier.

I would assume that on complex architectures with write buffers and
out-of-order prefetching, those are required, while on architectures
without those features, the barriers are cheap.

Arnd