Commit 158fd3ce authored by Peter Portante's avatar Peter Portante Committed by Anthony Liguori

qemu-timer.c: Remove 250us timeouts

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.
Signed-off-by: default avatarPeter Portante <>
Signed-off-by: default avatarAnthony Liguori <>
parent fc34e77b
......@@ -472,16 +472,16 @@ void qemu_run_all_timers(void)
alarm_timer->pending = 0;
/* vm time timers */
/* rearm timer, if not periodic */
if (alarm_timer->expired) {
alarm_timer->expired = 0;
/* vm time timers */
#ifdef _WIN32
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