linux-kernel - Re: [PATCH v2 1/3] net: irda: pxaficp_ir: use sched

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Message-ID: <87si6br8ev.fsf@belgarion.home>
Date:	Fri, 18 Sep 2015 18:36:56 +0200
From:	Robert Jarzmik <robert.jarzmik@...e.fr>
To:	David Miller <davem@...emloft.net>
Cc:	samuel@...tiz.org, petr.cvek@....cz, netdev@...r.kernel.org,
	linux-kernel@...r.kernel.org, arnd@...db.de
Subject: Re: [PATCH v2 1/3] net: irda: pxaficp_ir: use sched_clock() for time management

David Miller <davem@...emloft.net> writes:

>> My understanding is that the flow will be :
>>  sched_clock()
>>    rd->read_sched_clock() (cyc_to_ns() transformed for return)
>>      pxa_read_sched_clock()
>>        readl_relaxed(OSCR)
>> 
>> I didn't see any timings issue, as the flow looks equivalent to the readl(OSCR),
>> but I might have overlooked something.
>
> Of course it's different, because sched_clock() converts the value read
> from OSCR into nanoseconds, which is obviously different from using the
> OSCR register value directly.
>
> You're therefore feeding different values into this IRDA code.
Ah yes, I see it.
Which brings me to wonder which is the more correct :
 (a) replace to reproduce the same calculation
     Previously mtt was compared to a difference of 76ns steps (as 307ns / 4 =
     76ns):
     while ((sched_clock() - si->last_clk) * 76 < mtt)

 (b) change the calculation assuming mtt is in microseconds :
     while ((sched_clock() - si->last_clk) * 1000 < mtt)

I have no IRDA protocol knowledge so unless someone points me to the correct
calculation I'll try my luck with (b).

Cheers.

-- 
Robert
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