Messages in this thread | | | Subject | Re: [PATCH] drm: Fix lock order reversal between mmap_sem and struct_mutex. | From | Peter Zijlstra <> | Date | Fri, 20 Feb 2009 09:47:22 +0100 |
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On Fri, 2009-02-20 at 09:31 +0100, Thomas Hellstrom wrote: > Peter Zijlstra wrote: > > On Thu, 2009-02-19 at 22:02 +0100, Thomas Hellstrom wrote: > > > >> > >> It looks to me like the driver preferred locking order is > >> > >> object_mutex (which happens to be the device global struct_mutex) > >> mmap_sem > >> offset_mutex. > >> > >> So if one could avoid using the struct_mutex for object bookkeeping (A > >> separate lock) then > >> vm_open() and vm_close() would adhere to that locking order as well, > >> simply by not taking the struct_mutex at all. > >> > >> So only fault() remains, in which that locking order is reversed. > >> Personally I think the trylock ->reschedule->retry method with proper > >> commenting is a good solution. It will be the _only_ place where locking > >> order is reversed and it is done in a deadlock-safe manner. Note that > >> fault() doesn't really fail, but requests a retry from user-space with > >> rescheduling to give the process holding the struct_mutex time to > >> release it. > >> > > > > It doesn't do the reschedule -- need_resched() will check if the current > > task was marked to be scheduled away,
> Yes. my mistake. set_tsk_need_resched() would be the proper call. If I'm > correctly informed, that would kick in the scheduler _after_ the > mmap_sem() is released, just before returning to user-space.
Yes, but it would still life-lock in the RT example given in the other email.
> > furthermore yield based locking > > sucks chunks. > > > Yes, but AFAICT in this situation it is the only way to reverse locking > order in a deadlock safe manner. If there is a lot of contention it will > eat cpu. Unfortunately since the struct_mutex is such a wide lock there > will probably be contention in some situations.
I'd be surprised if this were the only solution. Maybe its the easiest, but not one I'll support.
> BTW isn't this quite common in distributed resource management, when you > can't ensure that all requestors will request resources in the same order? > Try to grab all resources you need for an operation. If you fail to get > one, release the resources you already have, sleep waiting for the > failing one to be available and then retry.
Not if you're building deterministic systems. Such constructs are highly non-deterministic.
Furthermore, this isn't really a distributed system is it?
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