[Intel-gfx] [PATCHv2] lib/ratelimit: Lockless ratelimiting
Dmitry Safonov
dima at arista.com
Tue Jun 26 17:46:23 UTC 2018
Hi Andy, thanks for the review,
On Tue, 2018-06-26 at 20:04 +0300, Andy Shevchenko wrote
[..]
> > #define RATELIMIT_STATE_INIT(name, interval_init, burst_init)
> > { \
> > - .lock =
> > __RAW_SPIN_LOCK_UNLOCKED(name.lock), \
>
> name is now redundant, isn't it?
It is. Worth to split on the second patch or keep callers changes in
this patch?
> > @@ -42,9 +41,10 @@ static inline void ratelimit_state_init(struct
> > ratelimit_state *rs,
> > {
> > memset(rs, 0, sizeof(*rs));
> >
> > - raw_spin_lock_init(&rs->lock);
> > rs->interval = interval;
> > rs->burst = burst;
> > + atomic_set(&rs->printed, 0);
> > + atomic_set(&rs->missed, 0);
>
> Can it be
>
> *rs = RATELIMIT_STATE_INIT(interval, burst);
>
> ?
>
> (Yes, the '(struct ratelimit_state)' has to be added to macro to
> allow this)
Sure.
> > static inline void ratelimit_state_exit(struct ratelimit_state
> > *rs)
> > {
> > + int missed;
> > +
> > if (!(rs->flags & RATELIMIT_MSG_ON_RELEASE))
> > return;
> >
> > - if (rs->missed) {
> > + if ((missed = atomic_xchg(&rs->missed, 0)))
>
> Perhaps
>
> missed = ...
> if (missed)
>
> ?
Ok, will change - checkpatch has warned me, but I thought it's just a
preference than a rule.
>
> > pr_warn("%s: %d output lines suppressed due to
> > ratelimiting\n",
> > - current->comm, rs->missed);
> > - rs->missed = 0;
> > - }
> > + current->comm, missed);
> > }
> > +static void ratelimit_end_interval(struct ratelimit_state *rs,
> > const char *func)
> > +{
> > + rs->begin = jiffies;
> > +
> > + if (!(rs->flags & RATELIMIT_MSG_ON_RELEASE)) {
> > + unsigned missed = (unsigned)atomic_xchg(&rs-
> > >missed, 0);
> > +
> > + if (missed)
> > + pr_warn("%s: %u callbacks suppressed\n",
> > func, missed);
>
> Instead of casting, perhaps
>
> int missed = ...
>
> I think you already has a guard against going it below zero. Or I
> missed something?
No, I do:
atomic_add_unless(&rs->missed, 1, -1);
So, it's guard against overflow, but not against negative.
That's why I do print it as unsigned.
--
Thanks,
Dmitry
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