Hi Xianda, Easiest answer first:
2. You need to write a forecast if, and only if, you're using general_work. I generally try to avoid doing that. Then: 1. ninput_items_required is, as you can see in the function signature, a reference to a vector. The size of the vector is the number of input ports. Compare to http://gnuradio.org/doc/doxygen/classgr_1_1block.html#a5bc118d94944d2ff71e378f807fb8d28 Greetings, Marcus On 04.06.2014 08:22, xianda wrote: > Hi all: > I want to know something about the forecast(). > I have already known that forcast() can tell scheduler how many > input items are required for each output item. > 1.But now i have read two example: > The first one: > void your_block::forecast(int > noutput_items,gr_vector_int &ninput_items_required){ > ninput_items_required[0]=100 * > noutput_items; > > ninput_items_required[1]=100 * > noutput_items; } > I have already understand it. > But the second one: > void forecast (int noutput_items, gr_vector_int > &ninput_items_required) > { > unsigned ninputs = ninput_items_required.size > (); > for (unsigned i = 0; i < ninputs; i++) > ninput_items_required[i] = 1;} > I can't understand since we can't know how many input > items we required,why use ninput_items_required.size ().Can someone help me? > 2.I want to know if we use the general_work().Is it means that > we must use the forcast()?Thanks. > Best regards > > > > > > _______________________________________________ > Discuss-gnuradio mailing list > Discuss-gnuradio@gnu.org > https://lists.gnu.org/mailman/listinfo/discuss-gnuradio
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