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>

drivers/net/ethernet/intel/ice/ice_txrx.c

@@ -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;