On 11/29/2018 01:43 PM, Davidlohr Bueso wrote: > On Thu, 29 Nov 2018, Peter Zijlstra wrote: > >> On Thu, Nov 29, 2018 at 02:12:32PM +0100, Peter Zijlstra wrote: >>> Yes, I think this is real, and worse, I think we need to go audit all >>> wake_q_add() users and document this behaviour. >>> >>> In the ideal case we'd delay the actual wakeup to the last wake_up_q(), >>> but I don't think we can easily fix that. >> >> See commit: 1d0dcb3ad9d3 ("futex: Implement lockless wakeups"), I think >> that introduces the exact same bug. >> >> Something like the below perhaps, altough this pattern seems to want a >> wake_a_add() variant that already assumes get_task_struct(). > > So I was looking at ways to avoid the redundant reference counting, > but given how wake_q_add() and wake_up_q() are so loose I can't > see how to avoid it -- we hold reference across the calls to maintain > valid mem. > > For example, wake_q will grab reference iff the cmpxchg succeeds, > likewise it will enter the wakeup loop in wake_up_q(), and there is no > awareness of which caller had the failed cmpxchg because another wakup > was in progress. >
I think it will be useful to make wake_q_add() to return the status of wake_q insertion so that we can respond appropriately if it fails. > And yes, afaict all wake_q users suffer from the same issue, so we have > to move the wake_q_add() after the condition, while explicitly doing > the task ref counting. How about adding two helpers - wake_q_enter(), wake_q_exit() that can hide all the reference counting? Thanks, Longman