Re: [libvirt] [PATCH 2/2] qemu: Change the default unix monitor timeout

On Thu, Jan 09, 2014 at 09:22:06AM +0100, Martin Kletzander wrote: > There is a number of reported issues when we fail starting a domain. > Turns out that, in some scenarios like high load, 3 second timeout is > not enough for qemu to start up to the phase where the socket is > created. Since the timeout is configurable and there is no downside > of waiting longer, raise the timeout right to 30 seconds. > > Signed-off-by: Martin Kletzander <mkletzan(a)redhat.com&gt; > --- > src/qemu/qemu.conf | 2 +- > src/qemu/qemu_monitor.c | 2 +- > 2 files changed, 2 insertions(+), 2 deletions(-) > > diff --git a/src/qemu/qemu.conf b/src/qemu/qemu.conf > index 6217b49..4936d88 100644 > --- a/src/qemu/qemu.conf > +++ b/src/qemu/qemu.conf > @@ -472,6 +472,6 @@ > # such file or directory" that could be because libvirt did not wait > # enough time, you can try increasing this timeout. > # > -# Default is 3 > +# Default is 30 > # > #monitor_socket_open_timeout = 60 > diff --git a/src/qemu/qemu_monitor.c b/src/qemu/qemu_monitor.c > index f34527a..6a437b1 100644 > --- a/src/qemu/qemu_monitor.c > +++ b/src/qemu/qemu_monitor.c > @@ -273,7 +273,7 @@ qemuMonitorOpenUnix(const char *monitor, pid_t cpid, virQEMUDriverPtr driver) > virQEMUDriverConfigPtr cfg = NULL; > struct sockaddr_un addr; > int monfd; > - int timeout = 3; /* In seconds */ > + int timeout = 30; /* In seconds */ > int ret; > size_t i = 0; ACK. It is safe to wait longer, since in the loop we kill() to check if QEMU is still running or not.