Subject: Re: [PATCH net] hv_netvsc: Fix napi reschedule while receive completion is busy

From: Stephen Hemminger <[email-masked]>
Date: Mon, 9 Jul 2018 11:15:27 -0700

On Mon,  9 Jul 2018 16:43:19 +0000
Haiyang Zhang <[email-masked]> wrote:

> From: Haiyang Zhang <[email-masked]>
> 
> If out ring is full temporarily and receive completion cannot go out,
> we may still need to reschedule napi if other conditions are met.
> Otherwise the napi poll might be stopped forever, and cause network
> disconnect.
> 
> Fixes: 7426b1a51803 ("netvsc: optimize receive completions")
> Signed-off-by: Haiyang Zhang <[email-masked]>
> ---
>  drivers/net/hyperv/netvsc.c | 8 ++++----
>  1 file changed, 4 insertions(+), 4 deletions(-)
> 
> diff --git a/drivers/net/hyperv/netvsc.c b/drivers/net/hyperv/netvsc.c
> index 8e9d0ee1572b..caaf5054f446 100644
> --- a/drivers/net/hyperv/netvsc.c
> +++ b/drivers/net/hyperv/netvsc.c
> @@ -1285,14 +1285,14 @@ int netvsc_poll(struct napi_struct *napi, int budget)
>  		nvchan->desc = hv_pkt_iter_next(channel, nvchan->desc);
>  	}
>  
> -	/* If send of pending receive completions suceeded
> -	 *   and did not exhaust NAPI budget this time
> +	send_recv_completions(ndev, net_device, nvchan);
> +
> +	/* If it did not exhaust NAPI budget this time
>  	 *   and not doing busy poll
>  	 * then re-enable host interrupts
>  	 *     and reschedule if ring is not empty.
>  	 */
> -	if (send_recv_completions(ndev, net_device, nvchan) == 0 &&
> -	    work_done < budget &&
> +	if (work_done < budget &&
>  	    napi_complete_done(napi, work_done) &&
>  	    hv_end_read(&channel->inbound) &&
>  	    napi_schedule_prep(napi)) {

This patch doesn't look right. I think the existing code works
as written.

If send_receive_completions is unable to send because ring is full
then vmbus_sendpacket will return -EBUSY which gets returns
from send_receive_completions.  Because the return is non-zero,
the driver will not call napi_complete_done.
Since napi_complete_done was not called, NAPI will reschedule
the napi poll routine.