Re: [PATCH 6/9] signal: Always call do_notify_parent_cldstop with siglock held

[Eric W. Biederman]

"Eric W. Biederman" <ebiederm@xxxxxxxxxxxx> writes:

> Oleg Nesterov <oleg@xxxxxxxxxx> writes:
>
>> On 04/26, Eric W. Biederman wrote:
>>>
>>> @@ -2164,7 +2166,9 @@ static void do_notify_parent_cldstop(struct task_struct *tsk,
>>>   	}
>>>
>>>  	sighand = parent->sighand;
>>> -	spin_lock_irqsave(&sighand->siglock, flags);
>>> +	lock = tsk->sighand != sighand;
>>> +	if (lock)
>>> +		spin_lock_nested(&sighand->siglock, SINGLE_DEPTH_NESTING);
>>
>> But why is it safe?
>>
>> Suppose we have two tasks, they both trace each other, both call
>> ptrace_stop() at the same time. Of course this is ugly, they both
>> will block.
>>
>> But with this patch in this case we have the trivial ABBA deadlock,
>> no?
>
> I was thinking in terms of the process tree (which is fine).
>
> The ptrace parental relationship definitely has the potential to be a
> graph with cycles.  Which as you point out is not fine.
>
>
> The result is very nice and I don't want to give it up.  I suspect
> something ptrace cycles are always a problem and can simply be
> forbidden.  That is going to take some analsysis and some additional
> code in ptrace_attach.
>
> I will go look at that.


Hmm.  If we have the following process tree.

    A
     \
      B
       \
        C

Process A, B, and C are all in the same process group.
Process A and B are setup to receive SIGCHILD when
their process stops.

Process C traces process A.

When a sigstop is delivered to the group we can have:

Process B takes siglock(B) siglock(A) to notify the real_parent
Process C takes siglock(C) siglock(B) to notify the real_parent
Process A takes siglock(A) siglock(C) to notify the tracer

If they all take their local lock at the same time there is
a deadlock.

I don't think the restriction that you can never ptrace anyone
up the process tree is going to fly.  So it looks like I am back to the
drawing board for this one.

Eric