Dear Marcus,
Sure, I have changed it, as follows:
void quadrator_upconverter_impl::forecast (int noutput_items,
gr_vector_int &ninput_items_required)
{ ninput_items_required[0] = std::min( 1, noutput_items / d_iFactor
); }
This tested has no effect to the flow graph; the block still doesn't
produce correct output when d_iFactor changes on the fly.
BTW I thought it is redundant to do this change, because as long as
set_min_noutput_items(d_iFactor) works on the fly, the scheduler will not
ask for less than iFactor samples. Correct..?
Regards,
Activecat
On Sat, Mar 8, 2014 at 5:41 PM, Marcus Müller <mar...@hostalia.de> wrote:
> Hi activecat,
> The scheduler might be confused if he asked for less than iFactor samples,
> because then your forecast tells him you need 0 input to produce that...
> The joys of Integer division
> Can you modify it so that it always demands at least 1 input?
>
> --
> Sent from my Android device with K-9 Mail. Please excuse my brevity.
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