Re: General question about TASK_INTERRUPTIBLE and schedule_timeout()

From: Shan Hai
Date: Fri Sep 02 2011 - 02:18:17 EST


On 09/01/2011 10:09 AM, Yong Zhang wrote:
On Wed, Aug 31, 2011 at 06:18:19PM +0530, sifram rajas wrote:
Hi,

I have a general question about the following 2 lines of code I see
all over the kernel:
1 set_current_state(TASK_INTERRUPTIBLE) ;
2 schedule_timeout(<some value>);

In the above code, if we encounter an interrupt after executing line
1, we will end up
call schedule() from the architecture specific code for CONFIG_PREEMPT
kernels, after
the interrupt handler has been invokled.
Yes.

This will cause the current task to sleep interruptibly forever

Actually, sleeping forever in the TASK_INTERRUPTIBLE state is not correct,
because even though the task is preempted by higher priority one
it will finally get a chance to run, but you will get time out value
of <some value> + preemption latency.

instead of for a certain timeout interval.
No.

schedule() will not put an preempted task to sleep, see:

This might be problematic, because on the IRQ to preemption check path
the PREEMPT_ACTIVE was already set and the following 'if' statement
could not hold because of
!(preempt_count() & PREEMPT_ACTIVE) == false

and the pick_next_task() might put the preempted task to sleep.

Correct me on any misunderstanding :-)

Cheers
Shan Hai

asmlinkage void __sched schduule(void)
{
...
if (prev->state&& !(preempt_count()& PREEMPT_ACTIVE)) {
if (unlikely(signal_pending_state(prev->state, prev))) {
prev->state = TASK_RUNNING;
} else {
...
}
}
...
}

Thanks,
Yong

Won't this defeat the purpose of the above code to schedule out or
sleep for a certain finite timeout ?
If yes, then what are the techniques to solve this problem ?


Thanks,
Sifram.
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