(Python) Source Block Outputs All 0's

Detlef_R · February 18, 2014, 7:51pm

Hi All,

I have implemented a python source block:

def __init__(self, kA1=4,kB1=4,k2=4,NA1=8,NB1=8,N2=8,M=8):
    gr.sync_block.__init__(self,
        name="blsd_enc_b",
        in_sig=None,
        out_sig=[numpy.uint8,numpy.uint8])

but I can’t get it to output anything but 0’s…

def work(self, input_items, output_items):
    out_cA = output_items[0]
    out_cB = output_items[1]

    out_cA = [ 5, 6, 7, 8, 9, 10 ] #cA[0]
    out_cB = [ 5, 6, 7, 8, 9, 10 ] #cB[0]
    self.produce(0,len(out_cA))
    self.produce(1,len(out_cB))
    print 'out_cA = ', out_cA, ', len(out_cA) = ', len(out_cA)
    print 'out_cB = ', out_cB, ', len(out_cB) = ', len(out_cB)
    return -1 #-2

I connect both outputs to a file sync of type byte, and it stores 6
bytes but its all “\00\00\00\00\00\00”

The print lines give:

out_cA = [5, 6, 7, 8, 9, 10] , len(out_cA) = 6
out_cB = [5, 6, 7, 8, 9, 10] , len(out_cB) = 6

as expected.

Any advice?

Regards,

David

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divotdave · February 18, 2014, 10:07pm

On Tue, Feb 18, 2014 at 12:50 PM, David Halls
[email protected] wrote:

print 'out_cA = ', out_cA, ', len(out_cA) = ', len(out_cA)
out_cB = [5, 6, 7, 8, 9, 10] , len(out_cB) = 6

as expected.

Any advice?

Regards,

David

David,

The problem here is that you’re not inserting items in to the out_cA
(and out_cB) arrays, you’re actually reassigning out_cA to be this new
list.

Honestly, I’ve never written a python block, but my understanding is
that you’d have to do something like this http://dpaste.com/1634971/

The items in the out_items list are actually numpy arrays, and they
need to be indexed. The [:] indexing is a shortcut of indexing the
entire array.

Another thing I’m not really sure about is whether the -1 will work or
not. I think you might have to return 6 so that the scheduler knows
there are 6 items to work on, and then the next time work is called
you return -1 ( assuming you’re wanting to stop the flowgraph here?).
Somebody else might chime in here if I’m wrong.

Nathan

divotdave · February 18, 2014, 11:13pm

Hi Nathan,

Thanks for your reply

I’ve done little python before, am trying to incorporate a collaborators
code into my system and he’s used python.

I am used to C, where it would copy the pointer. I understand your
advice - thanks, I will look at the link.

The return value is mimicking the WORK_CALLED_PRODUCE (-2) and WORK_DONE
return values that can be used when you have multiple outputs which may
output different number of items and the produce function is used.

I am not sure why the block keeps repeating if I use -2 as I use
successfully in many c++ blocks - I’m not used to writing source
blocks… maybe others have comments?