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

Mika Westerberg
2019-08-07T16:27:59Z

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? :-)