The scheduler chooses the amount of data to process each time. Here it chose
4 the first time it ran the general_work() function and the next time it
chose 1. You can look into the scheduler's implementation to know how it
chooses this. If you want to make it choose 5 the first time then you can
call gr_block::set_output_multiple() function to set the the output_items to
be a multiple of 5. This will make the scheduler choose 5 items on input
stream. There are other things involved too and I suggest looking at
gr_block.h file to know more.
2010/5/23 zzw.1012 <zzw.1...@163.com>
> Hi,
> Thanks to Seeve Bunch especially and sorry for my silly question. It's
> work well now. but , I still have some perplexity.
> here is the printf information in the file of howto_square_ff ::
> general_work():
>
> noutput_items = 4
> input_items[0] = -3.000000
> output_items[0] = 9.000000
> input_items[1] = 4.000000
> output_items[1] = 16.000000
> input_items[2] = -5.500000
> output_items[2] = 30.250000
> input_items[3] = 2.000000
> output_items[3] = 4.000000
> =======================
> noutput_items = 1
> input_items[0] = 3.000000
> output_items[0] = 9.000000
>
> while the source code is :
> howto_square_ff::general_work (int noutput_items,
> gr_vector_int &ninput_items,
> gr_vector_const_void_star &input_items,
> gr_vector_void_star &output_items)
> {
> const float *in = (const float *) input_items[0];
> float *out = (float *) output_items[0];
> //test
> *printf ("noutput_items = %d\n",noutput_items);*
> for (int i = 0; i < noutput_items; i++){
> *printf ("input_items[%d] = %f\n", i, in[i]);
> * out[i] = in[i] * in[i];
> *printf ("output_items[%d] = %f\n", i, out[i]);*
> }
> I can not understand that the src_data is src_data = (-3, 4, -5.5, 2, 3)
> which have 5 items in all, but why separate into 2 step ,the one is
> noutput_items = 4, the other is noutput_items = 1 ?
>
> 在2010-05-23，"Steve Bunch" <k9...@mac.com> 写道：
>
> const float *in = (const float *) input_items[0];
> float *out = (float *) output_items[0];
> ...
> printf ("input_items[%d] = %d\n", i, in[i]);
> out[i] = in[i] * in[i];
> printf ("output_items[%d] = %d\n", i, out[i]);
>
> in and out reference floating point numbers. You are printing them with %d
> instead of a floating point format, such as %f.
>
> Steve
> *
> *
>
>
>
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