Re: [PATCH net v5 1/3] net: sched: fix packet stuck problem for lockless qdisc

[Jakub Kicinski <kuba@kernel.org>]

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);
>  	}