Re: [patch V2 1/8] signal: Prevent exec() race
From: Frederic Weisbecker
Date: Mon Sep 07 2026 - 16:19:33 EST
Le Mon, Sep 07, 2026 at 05:26:04PM +0200, Thomas Gleixner a écrit :
> On Mon, Sep 07 2026 at 14:31, Frederic Weisbecker wrote:
> > Le Sat, Sep 05, 2026 at 08:59:01PM +0200, Thomas Gleixner a écrit :
> >> -out:
> >> - spin_unlock_irq(&tsk->sighand->siglock);
> >> + flush_sigqueue_list(&sigq_list);
> >
> > It probably doesn't matter in practice, I don't know feel free to ignore,
> > but FWIW it looks like it's still vulnerable to the theoretical far fetched
> > race I described. The head is moved under the lock but individual nodes are
> > deleted without the lock.
> >
> > CPU 0 CPU 1 CPU 2
> > ----- ----- -----
> >
> > exit_signals()
> > spin_lock(sighand)
> > tsk->flags |= PF_EXITING;
> > list_splice_init(&queue->list, head);
> > spin_unlock(sighand)
> >
> > list_for_each_safe(head, node)
> > list_del_init(node)
> > node->next = node // A
> > node->prev = node // B
> > ...
> > de_thread()
> > // acquired tsk->flags
> > // and signal flushed
> > // through tasklist_lock
> > transfer_pid() // C
> >
> > posix_timer_fn()
> > posixtimer_send_sigqueue()
> > // OBSERVES C
> > t = posixtimer_get_target(tmr)
> > lock_task_sighand()
> > // OBSERVES A
> > if (!list_empty(q))
> > // BUT NOT B
> > list_add_tail(q) // D
> >
> > Then who knows which write wins, B or D?
>
> For a moment you almost convinced me, but that's not possible:
>
> de_thread()
> ....
>
> if (!thread_leader()) {
> wait_until(old_leader->exit_state);
>
> transfer_pid();
>
> old_leader sets the exit_state in exit_notify():
>
> do_exit()
> exit_signals()
> lock(sighand)
> old_leader->flags |= PF_EXITING;
> head = remove_signals()
> unlock(sighand)
> flush_list(head)
> ...
> exit_notify()
> old_leader->exit_state = EXIT_XXX;
>
> From a program order POV the flush is completed _before_ the new leader
> can observe old_leader->exit_state and swap TIDS. exit_notify() and the
> wait in de_thread() are serialized via tasklist_lock.
Yes on that side all is program order. But the ordering is not mirrored on
the other side (at this stage of the patchset).
>
> The signal is either dropped before transfer_pid() is observable due to
> PF_EXITING on the old leader or queued on the new leader and then
> discarded in posixtimer_exit() -> flush_itimer_signals().
>
> The only valid question is whether it is guaranteed that on a weakly
> ordered system the stores in flush_sigqueue_list() are visible _before_
> transfer_pid() is visible to the third party.
>
> It's not obvious of course and might deserve a comment.
>
> exit_signals()
> lock(sighand)
> old_leader->flags |= PF_EXITING;
> head = remove_signals()
> #1 // RELEASE: PF_EXITING must become visible
> unlock(sighand)
> flush_list(head)
>
> ...
> posixtimer_exit()
> posix_cpu_timers_exit_task()
> lock(sighand)
> ...
> #2 // RELEASE: The stores in flush_list() must become visible
> // They might be already in case of preemption
> // or due a RELEASE operation in seccomp_filter_release()
> unlock(sighand)
That second step only appears at the end of the patchset, right? Otherwise
it's done on release_task(), which is after transfer_pid().
>
> ...
> exit_notify()
> lock(task_list_lock)
> exit_state = EXIT_ZOMBIE;
> #3 // RELEASE: exit_state must become visible
> unlock(task_list_lock)
>
> So the new leader cannot proceed before #3 which means it can't swap
> TIDs before that point. That requires task_list_lock so there is no way
> that the TID swap can trickle before the lock is held and exit_state
> being non-zero.
>
> Though the important part is that the third party on CPU3 has to acquire
> sighand lock in posixtimer_send_sigqueue(), which is an ACQUIRE
> operation. That means _all_ accesses to tsk::flags and to the sigqueue
> must happen _after_ the lock is acquired.
>
> If it acquires it after #1 and before the TID swap it must observe
> PF_EXITING and return immediately. So a concurrent modification of
> timer::sigqueue in flush_list() or not-yet visible stores are
> irrelevant.
>
> If it acquires it after #2 it must observe the full writes to the
> sigqueue. So after that point it does not longer matter whether the PID
> resolves to T1 or T2.
>
> No?
At the end of the patchset yes. But it doesn't look that way in this
very patch which is to be backported alone.
Thanks.
--
Frederic Weisbecker
SUSE Labs