There is a potential bug in "yield_to"
Hi, All
I found there may be a potential bug in "yield_to":
local_irq_save(flags);
rq = this_rq();again:
//task's rq may already changed in "sched_move_task"
p_rq = task_rq(p);
double_rq_lock(rq, p_rq);
while (task_rq(p) != p_rq) {
double_rq_unlock(rq, p_rq);
goto again;
}I think it may happen in this scene:
cpu 0 cpu 1(task a)
yield_to { disable_irq
sched_move_task { rq = this_rq();
task_rq_lock(task a) double_rq_lock
hold lock of rq 1
set_task_rq //task rq changed
release lock of rq 1
hold lock of rq 1 but task b no longer there
set rq 1's current to skip which is not task a
which means we hold a rq's lock but it's current is not the one should do yield.
Only "sched_move_task" will cause this issue as it will move the task which is still running.
The bug will make the task who want to do yield failed to set skip buddy to himself, but to a innocent task instead, not very harmful and almost impossible to occur in normal, but should we fix it with another check "rq == this_rq()"?
Regards,
Michael Wang
Quoted 41 lines · Michael Wang
Hi, All
I found there may be a potential bug in "
yield_to":local_irq_save(flags); rq = this_rq();again:
//task's rq may already changed in "
sched_move_task"p_rq = task_rq(p); double_rq_lock(rq, p_rq); while (task_rq(p) != p_rq) { double_rq_unlock(rq, p_rq); goto again; }I think it may happen in this scene:
cpu 0 cpu 1(task a)
yield_to{disable_irq
sched_move_task{ rq =this_rq();
task_rq_lock(task a)double_rq_lockhold lock of rq 1
set_task_rq//task rq changed
release lock of rq 1hold lock of rq 1 but task b no longer there
set rq 1's current to skip which is not task a
which means we hold a rq's lock but it's current is not the one should do yield.
Quoted 7 lines
Only "
sched_move_task" will cause this issue as it will move the task which is still running.The bug will make the task who want to do yield failed to set skip buddy to himself, but to a innocent task instead, not very harmful and almost impossible to occur in normal, but should we fix it with another check "rq ==
this_rq()"?
Uhm, what?!
We've got interrupts disabled, this_rq() cannot ever possibly change, so rq is always correct.
Only p_rq can change, and we have an again loop on that, so what's the problem again?
Quoted 50 lines · Peter Zijlstra
Quoted 41 lines · Michael Wang
Hi, All
I found there may be a potential bug in "
yield_to":local_irq_save(flags); rq = this_rq();again:
//task's rq may already changed in "
sched_move_task"p_rq = task_rq(p); double_rq_lock(rq, p_rq); while (task_rq(p) != p_rq) { double_rq_unlock(rq, p_rq); goto again; }I think it may happen in this scene:
cpu 0 cpu 1(task a)
yield_to{disable_irq
sched_move_task{ rq =this_rq();
task_rq_lock(task a)double_rq_lockhold lock of rq 1
set_task_rq//task rq changed
release lock of rq 1hold lock of rq 1 but task b no longer there
set rq 1's current to skip which is not task a
which means we hold a rq's lock but it's current is not the one should do yield.
Quoted 7 lines
Only "
sched_move_task" will cause this issue as it will move the task which is still running.The bug will make the task who want to do yield failed to set skip buddy to himself, but to a innocent task instead, not very harmful and almost impossible to occur in normal, but should we fix it with another check "rq ==
this_rq()"?
Quoted 5 lines
Uhm, what?!
We've got interrupts disabled,
this_rq()cannot ever possibly change, so rq is always correct.
I know I should have missed some thing, the schedule won't happen until enable the irq later, so even that scene happen, nothing will change on rq.
Thanks for your explain :)
Regards,
Michael Wang
Quoted 3 lines
Only
p_rqcan change, and we have an again loop on that, so what's the problem again?