No, do_tty_hangup() will never be called asynchronously: it will always be
called from a synchronous kernel context (either from within the scheduler
or from processes that do a run_task_queue() on their own).
If the caller already held the kernel lock, lock_kernel() is indeed a
no-op, but that's how it's designed: it doesn't need to do anything at
that point.
Basically, the problem is not the kernel lock itself - everything we do
runs with the kernel lock held (either gotten by do_tty_hangup(), or by
the entity that called it). The problem is that we cache a "tty" pointer
in between lock domains - so even though everything holds the kernel lock,
there is nothing that validates that the pointer is not stale.
Linus
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