Hi Bruce,
On 03/26/2015 10:14 PM, Bruce Richardson wrote:
> 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)) {
>
>
Acked-by: Olivier Matz <olivier.matz at 6wind.com>
Thanks!
