On Thu, 6 May 2021 09:57:42 +0800 Yunsheng Lin wrote:
> @@ -159,8 +160,37 @@ static inline bool qdisc_is_empty(const struct Qdisc *qdisc)
> static inline bool qdisc_run_begin(struct Qdisc *qdisc)
> {
> if (qdisc->flags & TCQ_F_NOLOCK) {
> + bool dont_retry = test_bit(__QDISC_STATE_MISSED,
> + &qdisc->state);
> +
> + if (spin_trylock(&qdisc->seqlock))
> + goto nolock_empty;
> +
> + /* If the flag is set before doing the spin_trylock() and
> + * the above spin_trylock() return false, it means other cpu
> + * holding the lock will do dequeuing for us, or it wil see
s/wil/will/
> + * the flag set after releasing lock and reschedule the
> + * net_tx_action() to do the dequeuing.
I don't understand why MISSED is checked before the trylock.
Could you explain why it can't be tested directly here?
> + */
> + if (dont_retry)
> + return false;
> +
> + /* We could do set_bit() before the first spin_trylock(),
> + * and avoid doing second spin_trylock() completely, then
> + * we could have multi cpus doing the set_bit(). Here use
> + * dont_retry to avoid doing the set_bit() and the second
> + * spin_trylock(), which has 5% performance improvement than
> + * doing the set_bit() before the first spin_trylock().
> + */
> + set_bit(__QDISC_STATE_MISSED, &qdisc->state);
> +
> + /* Retry again in case other CPU may not see the new flag
> + * after it releases the lock at the end of qdisc_run_end().
> + */
> if (!spin_trylock(&qdisc->seqlock))
> return false;
> +
> +nolock_empty:
> WRITE_ONCE(qdisc->empty, false);
> } else if (qdisc_is_running(qdisc)) {
> return false;
> @@ -176,8 +206,13 @@ static inline bool qdisc_run_begin(struct Qdisc *qdisc)
> static inline void qdisc_run_end(struct Qdisc *qdisc)
> {
> write_seqcount_end(&qdisc->running);
> - if (qdisc->flags & TCQ_F_NOLOCK)
> + if (qdisc->flags & TCQ_F_NOLOCK) {
> spin_unlock(&qdisc->seqlock);
> +
> + if (unlikely(test_bit(__QDISC_STATE_MISSED,
> + &qdisc->state)))
> + __netif_schedule(qdisc);
> + }
> }
>
> static inline bool qdisc_may_bulk(const struct Qdisc *qdisc)
> diff --git a/net/sched/sch_generic.c b/net/sched/sch_generic.c
> index 44991ea..9bc73ea 100644
> --- a/net/sched/sch_generic.c
> +++ b/net/sched/sch_generic.c
> @@ -640,8 +640,10 @@ static struct sk_buff *pfifo_fast_dequeue(struct Qdisc *qdisc)
> {
> struct pfifo_fast_priv *priv = qdisc_priv(qdisc);
> struct sk_buff *skb = NULL;
> + bool need_retry = true;
> int band;
>
> +retry:
> for (band = 0; band < PFIFO_FAST_BANDS && !skb; band++) {
> struct skb_array *q = band2list(priv, band);
>
> @@ -652,6 +654,16 @@ static struct sk_buff *pfifo_fast_dequeue(struct Qdisc *qdisc)
> }
> if (likely(skb)) {
> qdisc_update_stats_at_dequeue(qdisc, skb);
> + } else if (need_retry &&
> + test_and_clear_bit(__QDISC_STATE_MISSED,
> + &qdisc->state)) {
Why test_and_clear_bit() here? AFAICT this is the only place the bit
is cleared. So the test and clear do not have to be atomic.
To my limited understanding on x86 test_bit() is never a locked
operation, while test_and_clear_bit() is always locked. So we'd save
an atomic operation in un-contended case if we tested first and then
cleared.
> + /* do another dequeuing after clearing the flag to
> + * avoid calling __netif_schedule().
> + */
> + smp_mb__after_atomic();
test_and_clear_bit() which returned true implies a memory barrier,
AFAIU, so the barrier is not needed with the code as is. It will be
needed if we switch to test_bit() + clear_bit(), but please clarify
what it is paring with.
> + need_retry = false;
> +
> + goto retry;
> } else {
> WRITE_ONCE(qdisc->empty, true);
> }
