The below C program, run on my host, uses ~0% of the host CPU when run
with 100000 (e.g. wake up every 100us).
When run under kvm, this uses ~100% of the processor. The reason is there
is a vmexit for every timer wakeup.
Now, my host processor is E5-2699 v3, and this includes apicv
functionality. My host kernel is 4.9 (as is the guest). qemu is 2.5.
(ubuntu 16.04).
My understanding from the apicv feature was this timer could be entirely
done inside the guest on this generation (haswell) of processor.
In my application I'm finding that nearly all of my CPU(s) are going to
servicing vmexits from various threads that are waking up.
a) is there an efficient way to sleep/wakeup in a kvm guest?
b) is this what i should expect?
c) is there any work around or features i should be looking for?
Does anyone have any suggestions here?
#include <assert.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/timerfd.h>
int
main(int argc, char **argv)
{
int nr;
long long ns;
char buf[8];
struct itimerspec o, n;
if (argc < 2)
{
fprintf(stderr, "Usage: %s ns-to-wake\n", argv[0]);
exit(1);
}
ns = atoll(argv[1]);
printf("Wake up every %llu ns (%6.2fs)\n", ns, (1.0 * ns) / (1000
* 1000 * 1000));
int fd = timerfd_create(CLOCK_MONOTONIC, 0);
assert(fd > 0);
n.it_interval.tv_sec = 0;
n.it_interval.tv_nsec = 1;
n.it_value.tv_sec = (ns / (1000 * 1000 * 1000));
n.it_value.tv_nsec = ns % (1000 * 1000 * 1000);
do
{
assert(timerfd_settime(fd, 0, &n, &o) >= 0);
nr = read(fd, buf, sizeof(buf));
}
while (nr >= 0);
}
