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Re: [Discuss-gnuradio] How do the Scheduler and the flowgraph work?

From: John Andrews
Subject: Re: [Discuss-gnuradio] How do the Scheduler and the flowgraph work?
Date: Thu, 15 Jul 2010 17:24:07 -0700

This loop does all your work. Without it your block is useless. :)

To know how the scheduler chooses the value of items that are passed into the work() function I suggest you look into scheduler's code. It's a fairly easy to understand C++ code and most of your answers will be answered.

According to my knowledge the work() function always returns the number of items consumed by your block each time the work() function is called. When you return from work() you are basically informing the scheduler how many items were consumed so that the scheduler doesn't discard the unused input data. Believe me I once did the same mistake of not returning and had a troubling time knowing why my block wasn't working as it was supposed to be.

On Thu, Jul 15, 2010 at 3:02 AM, Sammour <address@hidden> wrote:

I have a basic question regarding GNURadio scheduler and how the flowgraph
works. I am using gr_sync_block but my question are valid for gr_block

>From the online tutorial about how to write a block I read:

 87    for (int i = 0; i < noutput_items; i++){
 88      out[i] = in[i] * in[i];
 89    }

I could print noutput_items and it is a huge number in order of thousands.
So my question is why should I create this loop? and how does it work? I
have read somewhere in this discution board that not using this loop will
increase the calclation overhead but no clue why. Does my block outputs the
items "bunch by bunch" or item by item.

Another question here: I saw few programs which dont return noutput_items
from work or general_work so it seems that I dont have to always return it.
How can I determine the value I shoud return.


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