Am Mittwoch, 31. Januar 2007 09:49 schrieb Rafael J. Wysocki: > On Wednesday, 31 January 2007 09:40, Oliver Neukum wrote: > > Am Mittwoch, 31. Januar 2007 09:33 schrieb Rafael J. Wysocki: > > > On Tuesday, 30 January 2007 23:32, Rafael J. Wysocki wrote: > > > > > > Generally, you are safe if your driver only calls wake_up() from a process > > > > context, but not from .resume() or .suspend() routines (or from an > > > > unfreezeable kernel thread). > > > > > > Ah, sorry, I've just realized I was wrong. Processes in TASK_UNINTERRUPTIBLE > > > cannot be frozen! So, the above only applies to wake_up_interruptible(). > > > > So the kernel will wait for tasks in TASK_UNINTERRUPTIBLE to finish IO > > before it calls suspend()? I am confused. > > Yes, it will. The process freezer can only return success if there are no more > TASK_UNINTERRUPTIBLE tasks. Otherwise it fails (after a timeout). So, this means, on suspend(): 1. Don't worry about TASK_UNINTERRUPTIBLE 2. Do worry about TASK_INTERRUPTIBLE We have to cease IO and must not call wake_up_interruptible() Isn't that a race until suspend() is called? On resume(): 1. Don't worry about TASK_UNINTERRUPTIBLE 2. Do not restart IO that may call wake_up_interruptible() When do we restart such IO? Regards Oliver
