[LTP] [PATCH v3 3/7] fzsync: Add long running thread support and deviation stats

[Richard Palethorpe]

Richard Palethorpe writes:

> Hello,
>
> Cyril Hrubis writes:
>
>> Hi!
>>> +/**
>>> + * tst_fzsync_pair_wait - Wait for the other thread
>>> + * @our_cntr: The counter for the thread we are on
>>> + * @other_cntr: The counter for the thread we are synchronising with
>>> + *
>>> + * Use this (through tst_fzsync_pair_wait_a() and tst_fzsync_pair_wait_b()) if
>>> + * you need an additional synchronisation point in a thread or you do not want
>>> + * to use the delay facility (not recommended). See
>>> + * tst_fzsync_pair_wait_update().
>>> + *
>>> + * Returns a non-zero value if the thread should continue otherwise the
>>> + * calling thread should exit.
>>> + */
>>> +static inline int tst_fzsync_pair_wait(struct tst_fzsync_pair *pair,
>>> +				       int *our_cntr, int *other_cntr)
>>> +{
>>> +	tst_atomic_inc(other_cntr);
>>> +	while (tst_atomic_load(our_cntr) < tst_atomic_load(other_cntr)
>>> +	       && !tst_atomic_load(&pair->exit))
>>> +		;
>>> +
>>> +	return !tst_atomic_load(&pair->exit);
>>> +}
>>
>> Also what happens if the counters overflow?
>>
>> I guess that this will happen if we call the test with large enough -i
>> paramter, right?
>
> Bad things will happen :-). It can probably be fixed with the following
> though:
>
>  static inline int tst_fzsync_pair_wait(struct tst_fzsync_pair *pair,
>  				       int *our_cntr, int *other_cntr)
>  {
> -	tst_atomic_inc(other_cntr);
> +	if (tst_atomic_inc(other_cntr) < 0) {
> +		tst_atomic_store(0, other_cntr);
> +		while (tst_atomic_load(our_cntr) > tst_atomic_load(other_cntr)
> +		       && !tst_atomic_load(&pair->exit))
> +			;
> +	}
> +
>  	while (tst_atomic_load(our_cntr) < tst_atomic_load(other_cntr)
>  	       && !tst_atomic_load(&pair->exit))
>
> Seems to work and any perf penality is insignificant; with or without
> branch prediction.

Actually this has a race condition if one thread is interrupted after
setting other_cntr to zero, but before it enters the while loop.

-- 
Thank you,
Richard.