On Tue, Jun 4, 2013 at 3:42 AM, Russell King - ARM Linux
<linux@xxxxxxxxxxxxxxxx> wrote:
> On Mon, Jun 03, 2013 at 02:11:59PM -0700, Stephen Boyd wrote:
>> On 06/03/13 02:39, Russell King - ARM Linux wrote:
>> > On Sat, Jun 01, 2013 at 11:39:41PM -0700, Stephen Boyd wrote:
>> >> +}
>> >> +
>> >> +void __init
>> >> +setup_sched_clock_64(u64 (*read)(void), int bits, unsigned long rate)
>> >> +{
>> >> + if (cd.rate > rate)
>> >> + return;
>> >> +
>> >> + BUG_ON(bits <= 32);
>> >> + WARN_ON(!irqs_disabled());
>> >> + read_sched_clock_64 = read;
>> >> + sched_clock_func = sched_clock_64;
>> >> + cd.rate = rate;
>> >> + cd.mult = NSEC_PER_SEC / rate;
>> > Here, you don't check that the (2^bits) * mult results in a wrap of the
>> > resulting 64-bit number, which is a _basic_ requirement for sched_clock
>> > (hence all the code for <=32bit clocks, otherwise we wouldn't need this
>> > complexity in the first place.)
>>
>> Ok I will use clocks_calc_mult_shift() here.
>
> No, that's not the problem.
>
> If you have a 56-bit clock which ticks at a period of 1ns, then
> cd.rate = 1, and your sched_clock() values will be truncated to 56-bits.
> The scheduler always _requires_ 64-bits from sched_clock. That's why we
> have the complicated code to extend the 32-bits-or-less to a _full_
> 64-bit value.
>
> Let me make this clearer: sched_clock() return values _must_ without
> exception monotonically increment from zero to 2^64-1 and then wrap
> back to zero. No other behaviour is acceptable for sched_clock().
Probably a trivial question.I was wondering why this particular requirement
exists in the first place.I looked into this commit 112f38a4a3 but couldn't
gather the reason.
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