rxq_cycle_sort is used to sort the rx queues by their measured number
of cycles. In the event that they are equal 0 could be returned.
However, it is observed that returning 0 results in a different sort
order on Windows/Linux. This is ok in practice but it causes a unit
test failure for
"1007: PMD - pmd-cpu-mask/distribution of rx queues" on Windows.
In order to have a consistent sort result, introduce a tiebreaker of
port/queue.
Fixes: 655856ef39b9 ("dpif-netdev: Change rxq_scheduling to use rxq processing
cycles.")
Reported-by: Alin Gabriel Serdean <[email protected]>
Signed-off-by: Kevin Traynor <[email protected]>
---
lib/dpif-netdev.c | 15 ++++++++++++---
1 file changed, 12 insertions(+), 3 deletions(-)
diff --git a/lib/dpif-netdev.c b/lib/dpif-netdev.c
index 0a62630..5627462 100644
--- a/lib/dpif-netdev.c
+++ b/lib/dpif-netdev.c
@@ -3452,4 +3452,5 @@ rxq_cycle_sort(const void *a, const void *b)
uint64_t total_qa, total_qb;
unsigned i;
+ int winner = -1;
qa = *(struct dp_netdev_rxq **) a;
@@ -3464,8 +3465,16 @@ rxq_cycle_sort(const void *a, const void *b)
dp_netdev_rxq_set_cycles(qb, RXQ_CYCLES_PROC_HIST, total_qb);
- if (total_qa >= total_qb) {
- return -1;
+ if (total_qa > total_qb) {
+ winner = 1;
+ } else if (total_qa == total_qb) {
+ /* Cycles are the same. Tiebreak on port/queue id. */
+ if (qa->port->port_no > qb->port->port_no) {
+ winner = 1;
+ } else if (qa->port->port_no == qb->port->port_no) {
+ winner = netdev_rxq_get_queue_id(qa->rx)
+ - netdev_rxq_get_queue_id(qb->rx);
+ }
}
- return 1;
+ return winner;
}
--
1.8.3.1
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