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

[Daniel P. Berrange]

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 at redhat.com>
> ---
>  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.


Daniel
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