Hi Nemanja,
Considering following flowgraph, assume your block is A, and assume all blocks work with the same itemsize. ![]() good question, but basically, when running, when A is done with a run of work, it's thread notifies blocks "upstream" (B in this case) that it has consumed ![]() ![]() In your case, this only happens for 0 output items; thus your general_work gets called with 0 input items, as Martin already stated. So what happens then? Since A has not provided any more output items for the blocks downstream, D can't do anything that it has not already been doing. The downstream part of the flowgraph stalls. Upstream part: Since GNU Radio has not been able to supply you with more than 0 input items, we can assume that the upstream part of the flowgraph is stuck, or done. Therefore, execution ends here. Greetings, Marcus On 10/18/2013 03:19 PM, Nemanja Savic wrote:
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