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mbuf: add explanation for confusing checks when freeing

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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>
This commit is contained in:
Bruce Richardson 2015-03-26 21:14:54 +00:00 committed by Thomas Monjalon
parent 2f95a470b8
commit 2911571232
1 changed files with 10 additions and 0 deletions
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10
lib/librte_mbuf/rte_mbuf.h
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@ -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)) {