Re: [External] [PATCH v2 4/8] riscv_cbqri: Add capacity controller probe and allocation device ops

[Drew Fustini <fustini@kernel.org>]

On Sat, Jun 27, 2026 at 05:31:03PM +0800, yunhui cui wrote:
> Hi Drew,
> 
> On Thu, Jun 25, 2026 at 9:41 AM Drew Fustini <fustini@kernel.org> wrote:
> >
> > Add support for the RISC-V CBQRI capacity controller. A platform driver
> > passes a cbqri_controller_info descriptor together with the cache level
> > to riscv_cbqri_register_cc_dt(), which probes the controller and adds it
> > to the controller list.
> >
> > Assisted-by: Claude:claude-opus-4-7
> > Co-developed-by: Adrien Ricciardi <aricciardi@baylibre.com>
> > Signed-off-by: Adrien Ricciardi <aricciardi@baylibre.com>
> > Signed-off-by: Drew Fustini <fustini@kernel.org>
[..]
> > diff --git a/drivers/resctrl/cbqri_devices.c b/drivers/resctrl/cbqri_devices.c
> > new file mode 100644
> > index 0000000000000000000000000000000000000000..8ad9df404f65d5d82722cf8b78f02936c489ca6d
> > --- /dev/null
> > +++ b/drivers/resctrl/cbqri_devices.c
[..]
> > +
> > +/* Set capacity block mask (cc_block_mask) */
> > +static void cbqri_set_cbm(struct cbqri_controller *ctrl, u64 cbm)
> > +{
> > +       iowrite64(cbm, ctrl->base + CBQRI_CC_BLOCK_MASK_OFF);
> 
> The CBQRI spec allows naturally aligned 4-byte accesses and only guarantees
> atomicity for 4-byte accesses; 8-byte atomicity is unspecified.
> 
> Would 32-bit split accesses be preferable here instead of relying on
> ioread64/iowrite64? This may also make the driver less dependent on native
> 64-bit MMIO support.

I suppose there could be systems that are RV64 but do 4-byte access for
the CBQRI registers. You are right the spec only guarantees atomicity
for naturally aligned 4-byte accesses and leaves 8-byte atomicity
unspecified.

I will switch the controller register accesses to 32-bit reads and
writes. The driver rejects ncblks > 32, so cc_block_mask only uses its
low 32 bits. For cc_alloc_ctl, the writable fields all sit in the low
word while the status and busy bits are read-only in the high word. A
read can reconstruct the value from two 32-bit reads and a write only
needs the low word. 

Thanks,
Drew