The logic used in the condition check before freeing an mbuf is
sometimes confusing, so explain it in a proper comment.
Signed-off-by: Bruce Richardson <bruce.richardson at intel.com>
---
lib/librte_mbuf/rte_mbuf.h | 10 ++++++++++
1 file changed, 10 insertions(+)
diff --git a/lib/librte_mbuf/rte_mbuf.h b/lib/librte_mbuf/rte_mbuf.h
index 17ba791..0265172 100644
--- a/lib/librte_mbuf/rte_mbuf.h
+++ b/lib/librte_mbuf/rte_mbuf.h
@@ -764,6 +764,16 @@ __rte_pktmbuf_prefree_seg(struct rte_mbuf *m)
{
__rte_mbuf_sanity_check(m, 0);
+ /*
+ * Check to see if this is the last reference to the mbuf.
+ * Note: the double check here is deliberate. If the ref_cnt is "atomic"
+ * the call to "refcnt_update" is a very expensive operation, so we
+ * don't want to call it in the case where we know we are the holder
+ * of the last reference to this mbuf i.e. ref_cnt == 1.
+ * If however, ref_cnt != 1, it's still possible that we may still be
+ * the final decrementer of the count, so we need to check that
+ * result also, to make sure the mbuf is freed properly.
+ */
if (likely (rte_mbuf_refcnt_read(m) == 1) ||
likely (rte_mbuf_refcnt_update(m, -1) == 0)) {
--
2.1.0
