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    SubjectRe: [PATCH] drm: Fix lock order reversal between mmap_sem and struct_mutex.
    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

    > > 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

    Furthermore, this isn't really a distributed system is it?

     \ /
      Last update: 2009-02-20 09:51    [W:0.032 / U:29.412 seconds]
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