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sched: fix grinder bug

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The rte_scheduler will get stuck and not deliver any more packets
if there are two active subports and then one of them stops enqueing
more packets. This is because of a bug in how the grinder state machines
are managed.

If a non-zero grinder is assigned (but not yet active), then the dequeue
would miss it and always return zero packets. The cure is to always
do a first pass over all grinders.

Signed-off-by: Stephen Hemminger <shemming@brocade.com>
Acked-by: Thomas Monjalon <thomas.monjalon@6wind.com>
This commit is contained in:
Stephen Hemminger 2014-05-14 09:25:27 -07:00 committed by Thomas Monjalon
parent dc05cabd98
commit 1f1f3a0078
1 changed files with 4 additions and 3 deletions
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7
lib/librte_sched/rte_sched.c
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@ -2113,12 +2113,12 @@ rte_sched_port_time_resync(struct rte_sched_port *port)
}
static inline int
rte_sched_port_exceptions(struct rte_sched_port *port)
rte_sched_port_exceptions(struct rte_sched_port *port, int second_pass)
{
int exceptions;
/* Check if any exception flag is set */
exceptions = (port->busy_grinders == 0) ||
exceptions = (second_pass && port->busy_grinders == 0) ||
(port->pipe_exhaustion == 1);
/* Clear exception flags */
@ -2140,7 +2140,8 @@ rte_sched_port_dequeue(struct rte_sched_port *port, struct rte_mbuf **pkts, uint
/* Take each queue in the grinder one step further */
for (i = 0, count = 0; ; i ++) {
count += grinder_handle(port, i & (RTE_SCHED_PORT_N_GRINDERS - 1));
if ((count == n_pkts) || rte_sched_port_exceptions(port)) {
if ((count == n_pkts) ||
rte_sched_port_exceptions(port, i >= RTE_SCHED_PORT_N_GRINDERS)) {
break;
}
}