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ice: Assume that more than one Rx queue is rare in ice_napi_poll 

Currently we divide budget by the number of Rx queues per Rx ring
container in ice_napi_poll even if there is only 1. This is an
unnecessary divide for the normal case of 1 Rx ring per Rx ring
container. Fix this by using an unlikely() call in the case where we
actually need to divide.

Also, we will always set budget_per_ring even if there are no Rx rings
in the Rx ring container so we don't need to initialize it to 0.

Signed-off-by: Brett Creeley <brett.creeley@intel.com>
Tested-by: Andrew Bowers <andrewx.bowers@intel.com>
Signed-off-by: Jeff Kirsher <jeffrey.t.kirsher@intel.com>
This commit is contained in:
Brett Creeley 2019-07-25 01:55:29 -07:00 committed by Jeff Kirsher
parent c1ddf1f5c4
commit 9118fcd525
1 changed files with 10 additions and 5 deletions
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15
drivers/net/ethernet/intel/ice/ice_txrx.c
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@ -1414,8 +1414,8 @@ int ice_napi_poll(struct napi_struct *napi, int budget)
container_of(napi, struct ice_q_vector, napi);
struct ice_vsi *vsi = q_vector->vsi;
bool clean_complete = true;
int budget_per_ring = 0;
struct ice_ring *ring;
int budget_per_ring;
int work_done = 0;
/* Since the actual Tx work is minimal, we can give the Tx a larger
@ -1429,11 +1429,16 @@ int ice_napi_poll(struct napi_struct *napi, int budget)
if (budget <= 0)
return budget;
/* We attempt to distribute budget to each Rx queue fairly, but don't
* allow the budget to go below 1 because that would exit polling early.
*/
if (q_vector->num_ring_rx)
/* normally we have 1 Rx ring per q_vector */
if (unlikely(q_vector->num_ring_rx > 1))
/* We attempt to distribute budget to each Rx queue fairly, but
* don't allow the budget to go below 1 because that would exit
* polling early.
*/
budget_per_ring = max(budget / q_vector->num_ring_rx, 1);
else
/* Max of 1 Rx ring in this q_vector so give it the budget */
budget_per_ring = budget;
ice_for_each_ring(ring, q_vector->rx) {
int cleaned;