ice: Assume that more than one Rx queue is rare in ice_napi_poll (9118fcd5) · Commits · 戴 / test

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

+10 −5

Original line number	Diff line number	Diff line
		@@ -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.
		/* 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.
		*/
		if (q_vector->num_ring_rx)
		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;

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