i40e: avoid double calling i40e_pf_rxq_wait()

Currently, when interface is being brought down and
i40e_vsi_stop_rings() is called, i40e_pf_rxq_wait() is called two times,
which is wrong. To showcase this scenario, simplified call stack looks
as follows:

i40e_vsi_stop_rings()
	i40e_control wait rx_q()
		i40e_control_rx_q()
		i40e_pf_rxq_wait()
	i40e_vsi_wait_queues_disabled()
		i40e_pf_rxq_wait()  // redundant call

To fix this, let us s/i40e_control_wait_rx_q/i40e_control_rx_q within
i40e_vsi_stop_rings().

Fixes: 65662a8d ("i40e: Fix logic of disabling queues")
Reviewed-by: Simon Horman <horms@kernel.org>
Signed-off-by: Maciej Fijalkowski <maciej.fijalkowski@intel.com>
Reviewed-by: Ivan Vecera <ivecera@redhat.com>
Tested-by: Chandan Kumar Rout <chandanx.rout@intel.com> (A Contingent Worker at Intel)
Signed-off-by: Tony Nguyen <anthony.l.nguyen@intel.com>

drivers/net/ethernet/intel/i40e/i40e_main.c

@@ -4926,7 +4926,7 @@ int i40e_vsi_start_rings(struct i40e_vsi *vsi)
 void i40e_vsi_stop_rings(struct i40e_vsi *vsi)
 {
 	struct i40e_pf *pf = vsi->back;
-	int pf_q, err, q_end;
+	int pf_q, q_end;
 
 	/* When port TX is suspended, don't wait */
 	if (test_bit(__I40E_PORT_SUSPENDED, vsi->back->state))
@@ -4936,16 +4936,10 @@ void i40e_vsi_stop_rings(struct i40e_vsi *vsi)
 	for (pf_q = vsi->base_queue; pf_q < q_end; pf_q++)
 		i40e_pre_tx_queue_cfg(&pf->hw, (u32)pf_q, false);
 
-	for (pf_q = vsi->base_queue; pf_q < q_end; pf_q++) {
-		err = i40e_control_wait_rx_q(pf, pf_q, false);
-		if (err)
-			dev_info(&pf->pdev->dev,
-				 "VSI seid %d Rx ring %d disable timeout\n",
-				 vsi->seid, pf_q);
-	}
+	for (pf_q = vsi->base_queue; pf_q < q_end; pf_q++)
+		i40e_control_rx_q(pf, pf_q, false);
 
 	msleep(I40E_DISABLE_TX_GAP_MSEC);
-	pf_q = vsi->base_queue;
 	for (pf_q = vsi->base_queue; pf_q < q_end; pf_q++)
 		wr32(&pf->hw, I40E_QTX_ENA(pf_q), 0);