Re: [PATCH 2/3] ptrace_stop: remove the wrong ->group_stop_count bookkeeping

Oleg Nesterov
2008-01-10T21:41:11Z

On 01/10, Petr Tesarik wrote:
>
> I can actually see a bug which may be related:
>
> 1. a process creates a thread (or more threads)
> 2. I attach/detach to that thread with strace several times
> (each time pressing CTRL-C to quit strace)
> 3. the whole thread group (except the traced thread) ends in
> TASK_STOPPED
>
> I looked at what strace was doing to that thread, and it sometimes sends
> SIGSTOP shortly before detaching. This is done when the thread is
> running, i.e. not waiting in ptrace_stop. Then PTRACE_DETACH returns
> - -ESRCH because it requires the tracee to be stopped -- just like all
> PTRACE_* requests except TRACEME and ATTACH. So, strace has no other
> option than to send an explicit SIGSTOP to the thread to stop it and
> discard it afterwards.
>
> Could this be related?

Perhaps yes. But there are so many oddities in this area. I don't know what
really happens with your test-case, but afaics this can happen even without
ptrace_stop() playing with the group stop.

Let's suppose that strace detached all sub-threads except T which is running,
and now strace does ptrace(PTRACE_DETACH, T). This fails, so strace does
kill(T, SIGSTOP).

Note that it use kill(), not tkill(). This means another sub-thread can
dequeue this signal and initiate the group stop (remember, it was already
detached and thus it is not traced any longer).

Now strace does wait4(T, __WALL). T notices the group stop in progress,
calls handle_group_stop(), and notifies its parent - strace.

wait4() returns success, strace does ptrace(PTRACE_DETACH, T) again. Now
T is TASK_STOPPED, ptrace() changes the state to TASK_TRACED and finally
does ptrace_untrace().

ptrace_untrace() sees TASK_TRACED. But it is possible that the group stop
is not completed yet (some sub-thread didn't pass handle_group_stop()), in
that case we are doing signal_wake_up(T, 1) so it becomes running.

I still think this series makes sense even if not complete.

Oleg.