RE: [PATCH 3/8] thunderbolt: Use 32-bit writes when writing ring producer/consumer

David Laight
2019-08-12T09:01:33Z

From: 'Mika Westerberg'
> Sent: 08 August 2019 10:58
> On Wed, Aug 07, 2019 at 04:41:30PM +0000, David Laight wrote:
> > From: 'Mika Westerberg' [mailto:mika.westerberg@linux.intel.com]
> > > Sent: 07 August 2019 17:36
> > >
> > > On Wed, Aug 07, 2019 at 04:22:26PM +0000, David Laight wrote:
> > > > From: Mika Westerberg
> > > > > Sent: 07 August 2019 17:14
> > > > > To: David Laight
> > > > >
> > > > > On Fri, Jul 05, 2019 at 04:04:19PM +0000, David Laight wrote:
> > > > > > > Really a matter of taste, but maybe you want to consider having a single
> > > > > > > function, with a 3rd parameter, bool is_tx.
> > > > > > > The calls here will be unified to:
> > > > > > > ring_iowrite(ring, ring->head, ring->is_tx);
> > > > > > > (No condition is needed here).
> > > > > > >
> > > > > > > The implementation uses the new parameter to decide which part of the register
> > > > > > > to mask, reducing the code duplication (in my eyes):
> > > > > > >
> > > > > > > val = ioread32(ring_desc_base(ring) + 8);
> > > > > > > if (is_tx) {
> > > > > > > val &= 0x0000ffff;
> > > > > > > val |= value << 16;
> > > > > > > } else {
> > > > > > > val &= 0xffff0000;
> > > > > > > val |= value;
> > > > > > > }
> > > > > > > iowrite32(val, ring_desc_base(ring) + 8);
> > > > > > >
> > > > > > > I'm not sure if it improves the readability or makes it worse. Your call.
> > > > > >
> > > > > > Gah, that is all horrid beyond belief.
> > > > > > If a 32bit write is valid then the hardware must not be updating
> > > > > > the other 16 bits.
> > > > > > In which case the driver knows what they should be.
> > > > > > So it can do a single 32bit write of the required value.
> > > > >
> > > > > I'm not entirely sure I understand what you say above. Can you shed some
> > > > > light on this by a concrete example how it should look like? :-)
> > > >
> > > > The driver must know both the tx and rx ring values, so:
> > > > iowrite32(tx_val << 16 | rx_val, ring_desc_base(ring) + 8);
> > > >
> > >
> > > I see. However, prod or cons side gets updated by the hardware as it
> > > processes buffers and other side is only updated by the driver. I'm not
> > > sure the above works here.
> >
> > If the hardware updates the other half of the 32bit word it doesn't ever work.
> >
> > In that case you must do 16bit writes.
> > If the hardware is ignoring the byte-enables it is broken and unusable.
>
> It is quite usable as I'm running this code on real hardware ;-) but
> 32-bit access is needed.
>
> The low or high 16-bits are read-only depending on the ring (Tx or Rx)
> and updated by the hardware. It ignores writes to that part so we could
> do this for Tx ring:
>
> iowrite32(prod << 16, ring_desc_base(ring) + 8);
>
> and this for Rx ring:
>
> iowrite32(cons, ring_desc_base(ring) + 8);
>
> Do you see any issues with this?

No, just comment that the hardware ignores the write to the other bytes.

You probably want separate functions - to remove the mispredicted branch.

David

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