On 2021-03-17 17:23:39 [+0100], Johan Hovold wrote:
> > > thread(irq_A)
> > >   irq_handler(A)
> > >     spin_lock(&foo->lock);
> > > 
> > > interrupt(irq_B)
> > >   irq_handler(B)
> > >     spin_lock(&foo->lock);
> > 
> > It will not because both threads will wake_up(thread).
> 
> Note that the above says "interrupt(irq_B)" suggesting it's a
> non-threaded interrupt unlike irq_A.

I missed that bit, thanks.

Sebastian

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