[PATCH] iommu/arm-smmu-v3: poll cmdq until it has space

[Sieber, Fernand]

Hi John,

> But is the polarity really correct? That is, if we don't have space,
> then exit with success (the function to check for space).

You are absolutely correct, this is a mistake that I made as I was resolving conflicts while porting this patch to iommu/next from 5.4 where I implemented and tested it.
It should be:

> -             if (!queue_full(llq))
> +             if (queue_has_space(llq, n))


> what is llq->state->val?

This is an other oversight for the same reason, llq->state->val has since then been renamed llq->val

Will fix both of these in the next revision.
Thanks and kind regards,

--Fernand

________________________________________
From: John Garry <john.garry at huawei.com>
Sent: Tuesday, September 21, 2021 18:22
To: Sieber, Fernand; will at kernel.org; robin.murphy at arm.com
Cc: linux-arm-kernel at lists.infradead.org; iommu at lists.linux-foundation.org; linux-kernel at vger.kernel.org
Subject: RE: [EXTERNAL] [PATCH] iommu/arm-smmu-v3: poll cmdq until it has space

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On 21/09/2021 12:43, Fernand Sieber wrote:
>       do {

I didn't follow the full logic of this change yet ...

>               llq->val = READ_ONCE(cmdq->q.llq.val);
> -             if (!queue_full(llq))
> +             if (!queue_has_space(llq, n))

But is the polarity really correct? That is, if we don't have space,
then exit with success (the function to check for space).

>                       break;
>
> +             /*
> +              * We must return here even if there's no space, because the producer
> +              * having moved forward could mean that the last thread observing the
> +              * SMMU progress has allocated space in the cmdq and moved on, leaving
> +              * us in this waiting loop with no other thread updating
> +              * llq->state->val.

what is llq->state->val?

> +              */
> +             if (llq->prod != prod)
> +                     return -EAGAIN;
> +
>               ret = queue_poll(&qp);

Thanks,
John