> >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.
> >From schedule():
> release_kernel_lock(prev, this_cpu);
The thing will _re-aquire_ the kernel lock when it comes through this
path. And then the lock_kernel() thing won't be a no-op like you said.
Sometimes it comes through other paths, and then lock_kernel _will_ be a
no-op, but then we will have the kernel lock anyway, so it doesn't matter.
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