[PATCH v20 08/17] clocksource/drivers/arm_arch_timer: Rework counter frequency detection.

On Thu, Jan 26, 2017 at 01:49:03PM +0800, Fu Wei wrote:

On 26 January 2017 at 01:25, Mark Rutland [off-list ref] wrote:

quoted

On Wed, Jan 25, 2017 at 02:46:12PM +0800, Fu Wei wrote:

quoted

On 25 January 2017 at 01:24, Mark Rutland [off-list ref] wrote:

quoted

On Wed, Jan 18, 2017 at 09:25:32PM +0800, fu.wei at linaro.org wrote:

quoted

From: Fu Wei <redacted>

quoted

For CNT{,EL0}BaseN.CNTFRQ, I am very concerned by the wording in the
current ARMv8 ARM ARM. This does not match my understanding, nor does it
match the description in the ARMv7 ARM. I believe this may be a
documentation error, and I'm chasing that up internally.

Either the currently logic in the driver which attempts to read
CNT{,EL0}BaseN.CNTFRQ is flawed, or the description in the ARM ARM is
erroneous.

Yes, those description did confuse me. :-(

But according to another document(ARMv8-A Foundation Platform User
Guide  ARM DUI0677K),
Table 3-2 ARMv8-A Foundation Platform memory map (continued)

AP_REFCLK CNTBase0, Generic Timer 64KB   S
AP_REFCLK CNTBase1, Generic Timer 64KB   S/NS

Dose it means the timer frame 0 can be accessed in SECURE status  only,
and the timer frame 1 can be accessed in both status?

That does appear to be what it says.

I assume in this case CNTCTLBase.CNTSAR<0> is RES0.

And because Linux kernel is running on Non-secure EL1, so should we
skip "SECURE" timer in Linux?

I guess you mean by checking the GTx Common flags, to see if the timer
is secure? Yes, we must skip those.

Looking further at this, the ACPI spec is sorely lacking any statement
as to the configuration of CNTCTLBase.{CNTSAR,CNTTIDR,CNTACR}, so it's
not clear if we can access anything in a frame, even if it is listed as
being a non-secure timer.

I think we need a stronger statement here. Otherwise, we will encounter
problems. Linux currently assumes that CNTCTLBase.CNTACR<N> is
writeable, given a non-secure frame N. This is only the case if
CNTCTLBase.CNTSAR.NS<N> == 1.

Thanks,
Mark.