I apologize if this sounds confused; it's because I am.
I'm trying to understand how thread dispatching works. Using the sound
example, I wrote an IRQ handler that does the following:
irq_handler()
{
clear_hardware_irq();
pulse_output();
rtl_hard_enable_irq(x);
}
Everything works fine. I look on my scope and can measure the time between
the input pulse that causes the interrupt and the output pulse from
pulse_output().
>From the same sound example, I then tried a variation where a thread is
dispatched from the handler, ie.
irq_handler()
{
pthread_wakeup_np(thread);
clear_hardware_irq();
rtl_hard_enable_irq(x);
}
and my thread looks like
void *my_thread(void *param)
{
rtl_stop_interrupts();
while (1) {
pthread_suspend_np(pthread_self());
pulse_output();
}
}
When I look on the scope I see 2 output pulses about 10usec apart, instead
of the expected 1 which I saw with the previous version. Since I don't have
a capture scope where I can freeze, I cannot say for certain there were
actually 2 pulses. But it sure looks this way. Only way for this to happen
is if the thread ran twice. Not possible, is it?
As a afternote, the system cannot be interrupted twice because the input
that generated the interrupt is vetoed until I do pulse_output() which
removes the veto. In any case the input rate is too slow.
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