On 04/30, Oleg Nesterov wrote:
pidfd_poll() can race with the exiting task, miss exit_code != 0, and return zero. However, do_poll() won't block after that and pidfd_poll() will be called again.
Here also I didn't follow what you mean. If exit_code is read as 0 in pidfd_poll(), then in do_poll() the count will be 0 and it will block in poll_schedule_timeout(). Right?
No. Please note the pwq->triggered check and please read __pollwake().
But if you want to understand this you can forget about poll/select. It is a bit complicated, in particular because it has to do set_current_state() right before schedule() and thus it plays games with pwq->triggered. But in essence this doesn't differ too much from the plain wait_event-like code (although you can also look at wait_woken/woken_wake_function).
If remove_wait_queue() could happem before wake_up_all() (like in your pseudo- code above), then pidfd_poll() or any other ->poll() method could miss _both_ the condition and wakeup. But sys_poll() doesn't do this, so it is fine to miss the condition and rely on wake_up_all() which ensures we won't block and the next iteration must see condition == T.
Oh, just in case... If it is not clear, of course I am talking about the case when wake_up_call() was already called when we check the condition. Otherwise everything is simple.
Oleg.