[dpdk-dev] i40e and memory allocations restricted on node 1
Ivan Nardi
nardi.ivan at gmail.com
Mon Feb 6 22:04:29 CET 2017
Hi
With dpdk 17.02-rc2 i40e driver doesn't load at all when memory
allocation is restricted on numa node 1 (on system with more than 1
node, obviously)
[root at micro ~]# /tmp/testpmd -d /home/micro/lib/librte_pmd_i40e.so -c
0xFFFC000 -w 0000:81:00.0 -w 0000:81:00.1 -n 4 --socket-mem=0,8192 --
-i --socket-num=1
EAL: Detected 56 lcore(s)
EAL: Probing VFIO support...
EAL: PCI device 0000:81:00.0 on NUMA socket 1
EAL: probe driver: 8086:1572 net_i40e
PMD: eth_i40e_dev_init(): FW 4.40 API 1.4 NVM 04.05.03 eetrack 80001cd8
RING: Cannot reserve memory
HASH: memory allocation failed
PMD: i40e_init_ethtype_filter_list(): Failed to create ethertype hash table!
EAL: Error - exiting with code: 1
Cause: Requested device 0000:81:00.0 cannot be used
Everything is fine with 16.11, or allowing allocations from both nodes
I was able to locate the code to fix (at least, I hope so), but I
don't really know if it is better to allocate from SOCKET_ID_ANY of
from the exact node associated with the network device, something like
dev->data->numa_node ("pseudo" patch below)
--- dpdk-17.02-rc2.orig/drivers/net/i40e/i40e_ethdev.c 2017-01-30
23:47:11.000000000 +0100
+++ dpdk-17.02-rc2/drivers/net/i40e/i40e_ethdev.c 2017-02-06
21:53:37.812313120 +0100
@@ -899,6 +899,7 @@
.entries = I40E_MAX_ETHERTYPE_FILTER_NUM,
.key_len = sizeof(struct i40e_ethertype_filter_input),
.hash_func = rte_hash_crc,
+ .socket_id = SOCKET_ID_ANY,
};
/* Initialize ethertype filter rule list and hash */
@@ -942,6 +943,7 @@
.entries = I40E_MAX_TUNNEL_FILTER_NUM,
.key_len = sizeof(struct i40e_tunnel_filter_input),
.hash_func = rte_hash_crc,
+ .socket_id = SOCKET_ID_ANY,
};
/* Initialize tunnel filter rule list and hash */
@@ -985,6 +987,7 @@
.entries = I40E_MAX_FDIR_FILTER_NUM,
.key_len = sizeof(struct rte_eth_fdir_input),
.hash_func = rte_hash_crc,
+ .socket_id = SOCKET_ID_ANY,
};
/* Initialize flow director filter rule list and hash */
Any thoughts?
Thanks in advance
Ivan
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