On Tue, Feb 18, 2003 at 09:44:19AM -0500, Chris Friesen wrote:
> > That certainly isn't what's supposed to happen. They should get woken
> > up in-order. The code in 2.4.18 seems to be doing that. Are you
> > doing anything clever with scheduling?
> static void locks_wake_up_blocks(struct file_lock *blocker,
> unsigned int wait)
> {
> while (!list_empty(&blocker->fl_block)) {
> struct file_lock *waiter = list_entry(blocker->fl_block.next,
> struct file_lock, fl_block);
> if (wait) {
> locks_notify_blocked(waiter);
>
> /* Let the blocked process remove waiter from the
> * block list when it gets scheduled.
> */
> current->policy |= SCHED_YIELD;
> schedule();
> } else {
> /* Remove waiter from the block list, because by the
> * time it wakes up blocker won't exist any more.
> */
> locks_delete_block(waiter);
> locks_notify_blocked(waiter);
> }
> }
> }
>
> It appears that if this function is called with a wait value of zero,
> all of the waiting processes will be woken up before the scheduler gets
> called. This means that the scheduler ends up picking which process
> runs rather than the locking code.
Right. That's why I asked whether you were doing something clever with
scheduling ;-)
> Looking through the file, there is no call chain on an unlock or on
> closing the last locked fd which can give a nonzero wait value, meaning
> that we will always end up with the scheduler making the decision in
> these cases.
I'm impressed that you chased it through ;-) This logic is mostly gone
from 2.5 because I found it too hard to keep in my mind while working
on this file.
> Am I missing something?
Nope, it's true. But the tasks get marked as runnable in the right order,
so the scheduler should be doing the right thing -- if any tasks really
have a better reason to run first (whether it's through RT scheduling
or through standard Unix priority scheduling) then they'll get the lock
first. Otherwise, I'd've thought it should be first-runnable, first-run.
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