Should we use "dsb" or "dmb" between write to buffer and write to register

[Will Deacon]

On Mon, Sep 12, 2022 at 07:16:26PM +0100, Catalin Marinas wrote:
> On Thu, Sep 08, 2022 at 02:50:17PM +0100, Will Deacon wrote:
> > On Wed, Sep 07, 2022 at 06:53:43PM +0100, Catalin Marinas wrote:
> > > On Mon, Aug 22, 2022 at 03:53:42PM +0800, Mark Zhang wrote:
> > > > So our questions are:
> > > > 1. can we use "dmb" here?
> > > > 2. If we can then should we use "dmb st", or "dmb oshst"?
> > > > 
> > > > Thank you very much.
> > > > 
> > > > [1] https://git.kernel.org/pub/scm/linux/kernel/git/rdma/rdma.git/commit/?id=22ec71615d824f4f11d38d0e55a88d8956b7e45f
> > > 
> > > Will convinced me at the time that it's sufficient, though every time I
> > > revisit this I get confused ;). Not sure whether we have updated the
> > > memory model since to cover such scenarios. In practice at least from
> > > what I recall that should be safe.
> > 
> > The Armv8 memory model is "other-multi-copy-atomic" which means that a
> > store is either visible _only_ to the observer from which it originates
> > or it is visible to all observers. It cannot exist in some intermediate
> > state.
> > 
> > With that, the insight is that a write to the MMIO interface of a shared
> > peripheral must be observed by all observers when it reaches the endpoint.
> 
> What's the endpoint here? The device itself or some serialisation point
> on the path to the device? IIUC, this can be a serialisation point in
> certain circumstances (e.g. with early write acknowledgement).

I don't think it matters too much, but you could replace "reaches the
endpoint" with "when the endpoint device begins to change state" if it
helps.

> > Consequently, we only need to ensure that the stores from your memcpy()
> > in the motivating example are observed before the MMIO write is observed
> > and a DMB ST is sufficient for that.
> 
> Yes but this is all about other observers observing the MMIO write
> rather than the device itself which cannot observe the MMIO write, so
> the CPU doesn't need to impose any order between these two.

Sorry, I don't understand you here. I agree that the device itself does not
observe the MMIO write, but the read transaction which it will generate in
response to the MMIO write is what we're trying to order against, and we need
to impose order between the memory writes and the MMIO write for that.

> Let's say we have a topology with two ports, one for MMIO and the other
> for RAM accesses, each with its own serialisation point:
> 
>   +-------+    +-------+
>   | CPU 0 |    | CPU 1 |
>   +-------+    +-------+
>     |   |        |   |
>    (a)--|--------+   |                  (a) MMIO serialisation point
>     |   +-----------(b)---+             (b) RAM serialisation point
>     |                |    |
>   +-----+        +-----+  |
>   | Dev |        | RAM |  |
>   +-----+        +-----+  |
>      |                    |
>      +-----DMA------------+
> 
> All accesses to RAM, including the device DMA, go through serialisation
> point (b). The MMIO accesses go through point (a). I don't know how
> realistic this is in practice (well, it can be a lot more complex) but
> with a few rules the above topology can obey the memory model. The
> simplest is for a DMB to cause the CPU to wait for the acknowledgement
> that a transaction reached a serialisation point before issuing new ones
> but there can be other ways like accesses issued on both ports before
> reaching the corresponding serialisation points. The serialisation
> points could communicate between them to ensure ordering in the presence
> of a third observer.
> 
> My worry is that in the absence of CPU1 (or transactions from CPU1), the
> hardware may decide to forward an MMIO access to the device even if it
> is ordered after a RAM transaction since it doesn't break any
> observability rules (it might as well consider the device private).

Right, I did distinguish between private and shared peripherals in my
initial response. A DSB may be required for private peripherals, but in
practice this is not the case and any such devices tend to have special
(arch-specific) drivers which can include the DSB (e.g. the GIC).

> > > I guess the question is what does it mean for the device that a third
> > > observer saw the write64. In one interpretation of observability,
> > > another write64 from the third observer is ordered after the original
> > > write64 but to me it still doesn't help clarify any order imposed on the
> > > device access to 'buff':
> > > 
> > > Initial state:
> > >   buff=0
> > >   ctrl=0
> > > 
> > > P0:		P1:		Device:
> > >   Wbuff=1	  Wctrl=2	  Ry=buff
> > >   DMB		  DMB
> > >   Wctrl=1	  Rx=buff
> > > 
> > > If the final 'ctrl' register value is 2 then x==1. But I don't see how
> > > y==0 or 1 is influenced by Wctrl=2. If x==1 on P1, any other observer,
> > > including the device, should see the buff value of 1 but this assumes
> > > that there is some other ordering for when Ry=buff is issued.
> > 
> > You need to relate the write to 'ctrl' with the device's read of 'buff'
> > somehow. Under which circumstances does the device read 'buff' (i.e.
> > what are the register fields in 'ctrl')?
> 
> I don't think we have anything in the memory model that can relate the
> write to MMIO with the device read from memory (DMA) since the device
> doesn't do a 'master' access to its own registers (i.e. go through
> serialisation point (a)). That's where I fail to explain in terms of the
> memory model why a DMB is sufficient (but I'm far from an expert here).

The architecture does actually have "out-of-band-ordered-before" for this
case, but it is built around DSB because it has to deal with the general
case which includes private peripherals.

> The scenario I have in mind is that P0 might forward the Wctrl=1 before
> Wbuff=1 reaches serialisation point (b) (e.g. there is some congestion
> on that port). If Wctlr=2 on P1 arrives at (a) after Wctrl=1,
> serialisation point (a) could stall it until point (b) confirms that all
> transactions prior to DMB have been sent so that the P0/P1 ordering is
> respected. However, this has no effect on the device observing Wbuff=1.
>
> I think the other way around holds - if the device observes Wbuff=1, the
> P1 must observe it as well. But if the device doesn't observe Wbuff=1,
> nothing breaks AFAICT.
> 
> I think we need a whiteboard (or a table in a pub after Plumbers).

Happy to. I can't tell from your diagram and text whether CPU1 can read
writes from CPU0 before they've reached (a) or (b).

Will