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