Re: [Qemu-devel] [PATCH] Small change to remove short 250us timeouts every other time through the wait loop.

[Avi Kivity <avi@redhat.com>]

On 03/22/2012 04:59 PM, Peter Portante wrote:
> Basically, the main wait loop calls qemu_run_all_timers() unconditionally. The
> first thing this routine used to do is to see if a timer had been serviced,
> and then reset the loop timeout to the next deadline.
>
> However, the new deadlines had not been calculated at that point, as
> qemu_run_timers() had not been called yet for each of the clocks. So
> qemu_rearm_alarm_timer() would end up with a negative or zero deadline, and
> default to setting a 250us timeout for the loop.
>
> As qemu_run_timers() is called for each clock, the real deadlines would be put
> in place, but because a loop timeout was already set, the loop timeout would
> not be changed.
>
> Once that 250us timeout fired, the real deadline would be used for the
> subsequent timeout.
>
> For idle VMs, this effectively doubles the number of times through the loop,
> doubling the number of select() system calls, timer calls, etc. putting added
> scheduling pressure on the kernel. And under cgroups, this really causes a big
> problem because the cgroup code does not scale well.
>
> By simply running the timers before trying to rearm the timer, we always rearm
> with a non-zero deadline, effectively halving the number of system calls.

Reviewed-by: Avi Kivity <avi@redhat.com>

Note the canonical subject line for patches is "subsystem: short
description", in this case something like "qemu-timer: remove spurious
host alarm wakeups" would be a good fit.

-- 
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