On 01/15/2019 02:57 PM, Zhirun Yan wrote:
> Before this patch, VF must request a specific queues(1/2/4/8/16) with
> DPDK PF. This patch align the number of requested queues to next power
> of 2. So VF can request any number queues from 1 to 16.
>
> Signed-off-by: Zhirun Yan <zhirun....@intel.com>
> ---
> drivers/net/i40e/i40e_pf.c | 5 ++++-
> 1 file changed, 4 insertions(+), 1 deletion(-)
>
> diff --git a/drivers/net/i40e/i40e_pf.c b/drivers/net/i40e/i40e_pf.c
> index 092e0d3e9..d110dffb7 100644
> --- a/drivers/net/i40e/i40e_pf.c
> +++ b/drivers/net/i40e/i40e_pf.c
> @@ -1271,7 +1271,10 @@ i40e_pf_host_process_cmd_request_queues(struct
> i40e_pf_vf *vf, uint8_t *msg)
> } else {
> i40e_vc_notify_vf_reset(vf);
> vf->vsi->nb_qps = req_pairs;
> - pf->vf_nb_qps = req_pairs;
> + if (rte_is_power_of_2(req_pairs))
> + pf->vf_nb_qps = req_pairs;
> + else
> + pf->vf_nb_qps = i40e_align_floor(req_pairs) << 1;
Shouldn't you do this before you validate against the number of free
queue pairs? or it is somehow handled
> i40e_pf_host_process_cmd_reset_vf(vf);
>
> return 0;
>
