Messages in this thread | | | Date | Fri, 18 Sep 2015 17:00:44 +0200 | From | Oleg Nesterov <> | Subject | Re: [PATCH] kernel: fix data race in put_pid |
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On 09/18, Peter Zijlstra wrote: > > On Fri, Sep 18, 2015 at 03:28:44PM +0200, Oleg Nesterov wrote: > > On 09/18, Peter Zijlstra wrote: > > > > > > ns = pid->numbers[pid->level].ns; > > > if ((atomic_read(&pid->count) == 1) || > > > atomic_dec_and_test(&pid->count)) { > > > > > > + smp_read_barrier_depends(); /* ctrl-dep */ > > > > Not sure... Firstly it is not clear what this barrier pairs with. And I > > have to admit that I can not understand if _CTRL() logic applies here. > > The same for atomic_read_ctrl(). > > The control dependency barrier pairs with the full barrier of > atomic_dec_and_test.
Yes thanks. I already got it. I hope ;)
> > OK, please forget about put_pid() for the moment. Suppose we have > > > > X = 1; > > synchronize_sched(); > > Y = 1; > > > > Or > > X = 1; > > call_rcu_sched( func => { Y = 1; } ); > > > > > > > > Now. In theory this this code is wrong: > > > > if (Y) { > > BUG_ON(X == 0); > > } > > > > But this is correct: > > > > if (Y) { > > rcu_read_lock_sched(); > > rcu_read_unlock_sched(); > > BUG_ON(X == 0); > > } > > > > So perhaps something like this > > > > /* > > * Comment to explain it is eq to read_lock + read_unlock, > > * in a sense that this guarantees a full barrier wrt to > > * the previous synchronize_sched(). > > */ > > #define rcu_read_barrier_sched() barrier() > > > > make sense? > > > > > > And again, I simply can't understand if this code > > > > if (READ_ONCE_CTRL(Y)) > > BUG_ON(X == 0); > > > > to me it does _not_ look correct in theory. > > So control dependencies provide a load-store barrier. Your examples > above rely on a load-load barrier; BUG_ON(X == 0) is a load.
Yes, yes...
What I tried to say is that we could fix it another way. And even look at this problem from another angle. No, it is not that I think it would be better in this particular case, but still...
put_pid() could do
if (atomic_read(&pid->count) == 1) {
rcu_read_lock(); rcu_read_unlock(); kmem_cache_free(pid); }
if we observe atomic_read() == 1, we know that we have at least one gp pass after all other writes to this memory (namely hlist_del_rcu() which removes it from rcu-list). Because we can see atomic_read() == 1 until delayed_put_pid() (called by RCU) drops its reference.
and perhaps this lock + unlock pair (which is nop at least for _sched) makes some sense in general...
Oleg.
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