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

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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.

> The ioread32() is likely to be very slow - you only want to do them
> if absolutely necessary.
> The speed of the iowrite32() doesn't matter (much) since it is almost
> certainly 'posted' and execution continues while the bus cycle is
> in progress.

OK thanks for the explanation.



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