Re: [PATCH] rework/fix is_single_threaded()

[Oleg Nesterov <oleg@redhat.com>]

On 04/16, Oleg Nesterov wrote:
>
> Suppose we have a process P which shares ->mm with "task" (the argument), so
> we should return "false".
>
> P does clone(CLONE_VM) and exits. rcu_read_lock() can't guarantee we will
> see the new task with the same ->mm. And without ->mmap_sem P can call
> exit_mm() and set P->mm = NULL.
>
> Hmm. But we can just add a barrier?
>
> 	bool is_single_threaded(struct task_struct *task)
> 	{
> 		struct mm_struct *mm = task->mm;
> 		struct task_struct *p, *t;
> 		bool ret;
>
> 		if (atomic_read(&task->signal->live) != 1)
> 			return false;
>
> 		if (atomic_read(&mm->mm_users) == 1)
> 			return true;
>
> 		ret = false;
> 		rcu_read_lock();
> 		for_each_process(p) {
> 			if (unlikely(p->flags & PF_KTHREAD))
> 				continue;
> 			if (unlikely(p == task->group_leader))
> 				continue;
>
> 			t = p;
> 			do {
> 				if (unlikely(t->mm == mm))
> 					goto found;
> 				if (likely(t->mm))
> 					break;
>
> 				/* !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
> 				   t->mm == NULL. Perhaps it had the same ->mm ?
> 				   If t has forked CLONE_VM task and called exit_mm(),
> 				   make sure next_thread() or for_each_process()->next_task()
> 				   will see it.
> 				   !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
> 				*/
> 				smp_rmb();

Sorry, forgot to mention...

But what if P does clone(CLONE_VM), exits, and for_each_process/while_each_thread
doesn't see it? IOW, what if we already see the result of list_del_rcu() ?

I think, in that case we must also see the result of clone()->list_add_tail_rcu()
which has a barrier, so we are safe.

Hmm. I feel this all has a simpler explanation, or I missed something...

Oleg.