Re: [PATCH] bpf: task_group_seq_get_next: cleanup the usage of next_thread()

From: Oleg Nesterov
Date: Mon Aug 21 2023 - 15:56:49 EST


So I still think the pid_alive() check should die...

and when I look at this code again I don't understand why does it abuse
task_struct->usage, I'll send another patch on top of this one.

On 08/21, Oleg Nesterov wrote:
>
> On 08/21, Kui-Feng Lee wrote:
> >
> >
> > On 8/21/23 08:09, Oleg Nesterov wrote:
> > >1. find_pid_ns() + get_pid_task() under rcu_read_lock() guarantees that we
> > > can safely iterate the task->thread_group list. Even if this task exits
> > > right after get_pid_task() (or goto retry) and pid_alive() returns 0 >
> > > Kill the unnecessary pid_alive() check.
> >
> > This function will return next_task holding a refcount, and release the
> > refcount until the next time calling the same function. Meanwhile,
> > the returned task A may be killed, and its next task B may be
> > killed after A as well, before calling this function again.
> > However, even task B is destroyed (free), A's next is still pointing to
> > task B. When this function is called again for the same iterator,
> > it doesn't promise that B is still there.
>
> Not sure I understand...
>
> OK, if we have a task pointer with incremented refcount and do not hold
> rcu lock, then yes, you can't remove the pid_alive() check in this code:
>
> rcu_read_lock();
> if (pid_alive(task))
> do_something(next_thread(task));
> rcu_read_unlock();
>
> because task and then task->next can exit and do call_rcu(delayed_put_task_struct)
> before we take rcu_read_lock().
>
> But if you do something like
>
> rcu_read_lock();
>
> task = find_task_in_some_rcu_protected_list();
> do_something(next_thread(task));
>
> rcu_read_unlock();
>
> then next_thread(task) should be safe without pid_alive().
>
> And iiuc task_group_seq_get_next() always does
>
> rcu_read_lock(); // the caller does lock/unlock
>
> task = get_pid_task(pid, PIDTYPE_PID);
> if (!task)
> return;
>
> next_task = next_thread(task);
>
> rcu_read_unlock();
>
> Yes, both task and task->next can exit right after get_pid_task(), but since
> can only happen after we took rcu_read_lock(), delayed_put_task_struct() can't
> be called until we drop rcu lock.
>
> What have I missed?
>
> Oleg.