Re: [Discuss-gnuradio] how to use FFT without grc block

[Marcus Müller]

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Hi Nasi,

Inbuf is the input buffer. It is a C++ pointer to a gr_complex.
That is the place where you put your input. As is described on the
doxygen page.

It is assumed that when you try to use a C++ framework, you are able
to understand basic C++ concepts. Again, I'd strongly agree with
Martin: This is using internal functionality of GNU Radio. You must be
able to read the source code to effectively employ it.

Sincerely,
Marcus

On 24.01.2014 15:55, Nasi wrote:
> And the problem is in that input part. It is not clear what is
> inbuf... I create gr_complex vector and want to input it into fft.
> It does not work in any way.  There are alot of questions are still
> open.
> 
> Coder is a good coder if his code is readable first. Anyone one can
> design a confusing language.
> 
> 
> Sent from Mail.Ru app for iOS
> 
> Freitag, 24. Januar 2014 13:59 +0100 from Martin Braun
> <address@hidden>: On 01/24/2014 02:45 AM, Nasi wrote:
>> Thanks!
>> 
>> with doxygen docs do you mean these:
>> http://gnuradio.org/doc/doxygen/classgr_1_1fft_1_1fft__complex.html
>> ? this redundant information is hopeless...
>> 
>> Do you know any normal good mature documentation?
> 
> Nasi,
> 
> part of learning GNU Radio is learning to read the documentation.
> You're pointing to a specific object deep inside the guts of GNU
> Radio. There will be no beginner-level documentation for these
> kinds of objects, probably ever. If you followed the docs through
> the navigation Modules -> Fourier Analysis, you'd see three blocks
> available for FFTs. All they do is calculate an FFT -- there is not
> much to say here. The assumption on this page is that you know how
> blocks work, and what an FFT is.
> 
> The page you pointed to is not redundant, whether or not it's
> hopeless is of course matter of debate. But it has all you need to
> calculate an FFT: The object you need, the functions you need to
> call etc. Here's all you really need: "compute FFT. The input comes
> from inbuf, the output is placed in outbuf."
> 
> Martin
> 
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