Messages in this thread | | | From | "Eric W. Biederman" <> | Date | Wed, 27 Apr 2022 09:47:10 -0500 | Subject | Re: [PATCH 6/9] signal: Always call do_notify_parent_cldstop with siglock held |
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"Eric W. Biederman" <ebiederm@xmission.com> writes:
> Oleg Nesterov <oleg@redhat.com> 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
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