Re: More parallel atomic_open/d_splice_alias fun with NFS and possibly more FSes.

[Oleg Drokin]

On Jul 5, 2016, at 1:42 PM, Al Viro wrote:

> On Tue, Jul 05, 2016 at 11:21:32AM -0400, Oleg Drokin wrote:
>>> ...
>>> -       if (d_unhashed(*de)) {
>>> +       if (d_in_lookup(*de)) {
>>>                struct dentry *alias;
>>> 
>>>                alias = ll_splice_alias(inode, *de);
>> 
>> This breaks Lustre because we now might progress further in this function
>> without calling into ll_splice_alias and that's the only place that we do
>> ll_d_init() that later code depends on so we violently crash next time
>> we call e.g. d_lustre_revalidate() further down that code.
> 
> Huh?  How the hell do those conditions differ there?

Like explained in my other email, because this is in a normal
lookup path, we can get here with a new dentry that was allocated in
__hash_lookup via d_alloc (not parallel) that's not marked with the PAR bit.

>> Also I still wonder what's to stop d_alloc_parallel() from returning
>> a hashed dentry with d_in_lookup() still true?
> 
> The fact that such dentries do not exist at any point?
> 
>> Certainly there's a big gap between hashing the dentry and dropping the PAR
>> bit in there that I imagine might allow __d_lookup_rcu() to pick it up
>> in between?--
> 
> WTF?  Where do you see that gap?  in-lookup dentries get hashed only in one
> place - __d_add().  And there (besides holding ->d_lock around both) we
> drop that bit in flags *before* _d_rehash().  AFAICS, the situation with
> barriers is OK there, due to lockref_get_not_dead() serving as ACQUIRE
> operation; I could be missing something subtle, but a wide gap...  Where?

Oh! I see, I missed that __d_add drops the PAR bit as well, not just the code
at the end of the call that does d_alloc_parallel.
Then indeed there is no gap, sorry for the false alarm.