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

David Laight
2019-08-07T16:41:35Z

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.

David

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