On 10/30, Balbir Singh wrote:
>
> Oleg Nesterov wrote:
>
> > 2. release_task(first) can happen after fill_tgid() drops tasklist_lock,
> > it is unsafe to dereference first->signal.
> >
>
> But, we have a reference to first via get_task_struct(). release_task()
> would do just a put_task_struct(). Am I missing something?
No, release_task() will reap the task. tsk->usage protects only task_struct
itself (more precisely, it protects against __put_task_struct()). And please
note that release_task()->__exit_signal() sets tsk->signal = NULL.
QUESTION: taskstats_exit_alloc() does kfree(kmem_cache_alloc()), is it OK?
Yes, it works, but is it good? The comment says:
* @objp: pointer returned by kmalloc.
Another question,
do_exit()
taskstats_exit_alloc()
...
taskstats_exit_send()
taskstats_exit_free()
What is the point? Why can't we have taskstats_exit() which does alloc+send+free
itself? This looks like unnecessary complication to me.
>From taskstats_exit_alloc:
/*
* This is the cpu on which the task is exiting currently and will
* be the one for which the exit event is sent, even if the cpu
* on which this function is running changes later.
*/
Why do we record current cpu exactly here? This task probably changed its
CPU many times since it entered sys_exit(), so what is the problem if it
will change CPU again before taskstats_exit_send() ?
Oleg.
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