mbuf: add explanation for confusing checks when freeing

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@intel.com>
Acked-by: Olivier Matz <olivier.matz@6wind.com>
Acked-by: Konstantin Ananyev <konstantin.ananyev@intel.com>
Acked-by: Neil Horman <nhorman@tuxdriver.com>

diff --git a/lib/librte_mbuf/rte_mbuf.h b/lib/librte_mbuf/rte_mbuf.h
index 17ba791..0265172 100644 (file)
--- 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)) {