Hello
The flowgraph of my top_block has test_demo as one of the blocks. The
work() function associated with it has only one print statement.
When I execute the flowgraph, the statement keeps on printing the statement
" I AM IN WORK" on the screen. Why does it happen?
My flowgraph is :
File source--->stream to datagram--->test_demo---->datagram to
stream----->file sink
test_demo.py is :
import timing
import numpy
import gras
import Queue
import thread
from gnuradio import gr
from gnuradio import digital
from gnuradio import uhd
import grextras
class test_demo(gras.Block):
def __init__(self, threshold):
gras.Block.__init__(self,name="test_demo",
in_sig = [numpy.uint8,numpy.uint8,numpy.uint8],
out_sig = [numpy.uint8,numpy.uint8])
self.input_config(0).reserve_items = 0
self.input_config(1).reserve_items = 0
self.output_config(1).reserve_items = 4096
self.output_config(0).reserve_items = 4096
self.threshold=threshold
print "inside test_demo"
def work(self,ins,outs):
print "I AM IN WORK"
If it is due to the size of buffer/queue of the block then how can I limit
it?
I understand that work() keeps on executing till the block receives
something at its input ports. I want work() to return the control to main
flowgraph upon satisfying a user specified condition such as
if (i=5):
self.mark_done()
It should ignore all the pending packets of the buffer/queue.
I want to start my flowgraph fresh after this process.
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