Messages in this thread | | | From | Thomas Gleixner <> | Subject | Re: [RFC PATCH] timer: Fix bucket_expiry calculation | Date | Wed, 12 May 2021 16:42:21 +0200 |
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Xiongfeng,
On Wed, May 12 2021 at 20:15, Xiongfeng Wang wrote: > When I use schedule_timeout(5) to put a process into sleep on my machine > with HZ = 100. It always sleep about 60ms. I enable the timer trace and > find out, when the timer_list expires, 'now' is always equal to > 'expires + 1'. I print 'base->next_expiry' in '__run_timers' and find out > 'next_expiry' is always equal to 'expires + 1'; > > It is because we use the following equation to calculate bucket_expiry. > > bucket_expiry = ((expires + LVL_GRAN(lvl)) >> LVL_SHIFT(lvl)) << LVL_SHIFT(lvl) > > 'bucket_expiry' is equal to 'expires + 1' when lvl = 0. So modify the > equation as follows to fix the issue. > > bucket_expiry = ((expires + LVL_GRAN(lvl) - 1) >> LVL_SHIFT(lvl)) << LVL_SHIFT(lvl)
That's wrong because you move the expiry of each timer one jiffie ahead, which violates the guarantee that a timer sleeps at least for one jiffie for real and not measured in jiffies.
jiffies = 0 schedule_timeout(1)
local_irq_disable() -> timer interrupt is raised in HW timer->expires = jiffies + 1 <- 1 add_timer(timer) local_irq_enable() timer interrupt jiffies++; softirq() expire(timer); -> timer is expired immediately
So the off by one has a reason and is required to prevent too short timeouts. There is nothing you can do about that because that's a property of low granularity tick based timer wheels.
That's even documented in the comment above the code you modified:
/* * The timer wheel has to guarantee that a timer does not fire * early. Early expiry can happen due to: * - Timer is armed at the edge of a tick * - Truncation of the expiry time in the outer wheel levels * * Round up with level granularity to prevent this. */
Thanks,
tglx
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