The RTE_ALIGN macro is aligned upwards. If the buf_size variable is not
aligned with 1 << I40E_RXQ_CTX_DBUFF_SHIFT, the rx_buf_len is larger than
the actual mbuf memory after the operation. When receiving the packet, if
the packet is larger than the configured buf_size, it will cause a memory
stepping event.
The patch uses the RTE_ALIGN_FLOOR down alignment macro to correct the
problem.
Signed-off-by: Qiming Chen <chenqiming_hua...@163.com>
---
drivers/net/i40e/i40e_ethdev_vf.c | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
diff --git a/drivers/net/i40e/i40e_ethdev_vf.c
b/drivers/net/i40e/i40e_ethdev_vf.c
index 924da8dfb4..5b1c8e76ab 100644
--- a/drivers/net/i40e/i40e_ethdev_vf.c
+++ b/drivers/net/i40e/i40e_ethdev_vf.c
@@ -1927,7 +1927,7 @@ i40evf_rxq_init(struct rte_eth_dev *dev, struct
i40e_rx_queue *rxq)
RTE_PKTMBUF_HEADROOM);
rxq->hs_mode = i40e_header_split_none;
rxq->rx_hdr_len = 0;
- rxq->rx_buf_len = RTE_ALIGN(buf_size, (1 << I40E_RXQ_CTX_DBUFF_SHIFT));
+ rxq->rx_buf_len = RTE_ALIGN_FLOOR(buf_size, (1 <<
I40E_RXQ_CTX_DBUFF_SHIFT));
len = rxq->rx_buf_len * I40E_MAX_CHAINED_RX_BUFFERS;
rxq->max_pkt_len = RTE_MIN(len,
dev_data->dev_conf.rxmode.max_rx_pkt_len);
--
2.30.1.windows.1
