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ice: Remove unnecessary wait when disabling/enabling Rx queues

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In ice_vsi_ctrl_rx_rings() we are unnecessarily waiting for
QRX_CTRL_QENA_REQ and QRX_CTRL_QENA_STAT to be the same value prior to
disabling each Rx queue. There is no reason to do this so remove
this wait loop as we already have a wait loop after disabling/enabling
the Rx queue through the QRX_CTRL register to make sure it gets
successfully disabled/enabled.

Signed-off-by: Brett Creeley <brett.creeley@intel.com>
Signed-off-by: Anirudh Venkataramanan <anirudh.venkataramanan@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 2019-02-28 15:25:56 -08:00 committed by Jeff Kirsher
parent b9c8bb06b5
commit acd1751a39
1 changed files with 2 additions and 8 deletions
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10
drivers/net/ethernet/intel/ice/ice_lib.c
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@ -197,19 +197,13 @@ static int ice_vsi_ctrl_rx_rings(struct ice_vsi *vsi, bool ena)
{
struct ice_pf *pf = vsi->back;
struct ice_hw *hw = &pf->hw;
int i, j, ret = 0;
int i, ret = 0;
for (i = 0; i < vsi->num_rxq; i++) {
int pf_q = vsi->rxq_map[i];
u32 rx_reg;
for (j = 0; j < ICE_Q_WAIT_MAX_RETRY; j++) {
rx_reg = rd32(hw, QRX_CTRL(pf_q));
if (((rx_reg >> QRX_CTRL_QENA_REQ_S) & 1) ==
((rx_reg >> QRX_CTRL_QENA_STAT_S) & 1))
break;
usleep_range(1000, 2000);
}
rx_reg = rd32(hw, QRX_CTRL(pf_q));
/* Skip if the queue is already in the requested state */
if (ena == !!(rx_reg & QRX_CTRL_QENA_STAT_M))