Con Kolivas wrote:
Convention in the kernel would be
aztTimeOut = HZ / 100 ? : 1;
to be at least one tick (works for HZ even below 100) and is at least 10ms. If
you wanted 2 ms then use
aztTimeOut = HZ / 500 ? : 1;
which would give you at least 2ms
Thank you Con for the feedback.
Hmm.. The minimum value should be 2, right?
Otherwise the loop could time out after only a few nanoseconds.. since the loop will then timeout immediately on a clock tick.
Or am I wrong?
Best regards,
Daniel Marjamäki
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