Re: [Discuss-gnuradio] something about forecast()

[Marcus Müller]

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

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