________________________________________ 发件人: Jens Axboe <axboe@xxxxxxxxx> 发送时间: 2020年10月27日 21:35 收件人: Zhang, Qiang 抄送: io-uring@xxxxxxxxxxxxxxx; linux-kernel@xxxxxxxxxxxxxxx 主题: Re: [PATCH] io-wq: set task TASK_INTERRUPTIBLE state before schedule_timeout On 10/26/20 9:09 PM, qiang.zhang@xxxxxxxxxxxxx wrote: > From: Zqiang <qiang.zhang@xxxxxxxxxxxxx> > > In 'io_wqe_worker' thread, if the work which in 'wqe->work_list' be > finished, the 'wqe->work_list' is empty, and after that the > '__io_worker_idle' func return false, the task state is TASK_RUNNING, > need to be set TASK_INTERRUPTIBLE before call schedule_timeout func. > >I don't think that's safe - what if someone added work right before you >call schedule_timeout_interruptible? Something ala: > > >io_wq_enqueue() > set_current_state(TASK_INTERRUPTIBLE(); > schedule_timeout(WORKER_IDLE_TIMEOUT); > >then we'll have work added and the task state set to running, but the >worker itself just sets us to non-running and will hence wait >WORKER_IDLE_TIMEOUT before the work is processed. > >The current situation will do one extra loop for this case, as the >schedule_timeout() just ends up being a nop and we go around again although the worker task state is running, due to the call schedule_timeout, the current worker still possible to be switched out. if set current worker task is no-running, the current worker be switched out, but the schedule will call io_wq_worker_sleeping func to wake up free worker task, if wqe->free_list is not empty. >checking for work. Since we already unused the mm, the next iteration >will go to sleep properly unless new work came in. > >-- >Jens Axboe
