Re: [Discuss-gnuradio] Assign to 'out' but don't 'produce' anything

[Richard Bell]

Ah OK. So the produce function advances the buffer pointer only?

Rich

On Wed, Jun 24, 2015 at 12:35 PM, Marcus Müller <address@hidden> wrote:

Hi Rich,

What happens when you assign input to output in a general_work call, out[ii]=in[ii], but don't call produce?

Then you're breaking a contract!
GNU Radio has to rely on you only writing samples you admit that you produce -- otherwise, the write pointer can't advance, and the next general_work will be offered the same buffer space again.

My minds eye sees the out variable as a secondary local buffer for general_work.

There is no local buffer! You directly work on the pseudocircular buffers; everything else would introduce unnecessary copy overhead.

Best regards,
Marcus

On 06/24/2015 09:23 PM, Richard Bell wrote:

What happens when you assign input to output in a general_work call, out[ii]=in[ii], but don't call produce? Is the stuff you dumped into 'out' lost when general_work returns WORK_CALLED_PRODUCE?

My minds eye sees the out variable as a secondary local buffer for general_work. You dump stuff in there while general_work has scope, and transfer the contents of this buffer to the gnuradio block buffer when you call produce. Am I understanding this correctly?

v/r,

Rich

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