Re: [PD] How latency works in cloned abstraction? (dsp graph order)

2022-01-14 Thread Christof Ressi
so if for example abstraction 0 sends to 1, that sends to 2, it would 
be 0 latency, but the opposite would be 128 samples?

With the current implementation: yes.

Note that the processing order of cloned instances is not really 
specified, so in theory you shouldn't rely on it. In practice, I don't 
see a reason why it would change.


Christof

On 15.01.2022 02:56, José de Abreu wrote:
Hello list, I have a curiosity, let's say that I want to clone an 
abstraction that reroutes signals between the cloned instances.


We learn that if a send~ is sorted before the corresponding receive~ 
we get no latency, but if the receive~ is sorted before, it reads from 
the previous block, so we get by default 64 samples of latency


But how does this order work in the case of cloned abstractions? the 
abstractions with a bigger or smaller "index" would be sorted 
accordingly (from 0 to n), or is it somewhat random?


so if for example abstraction 0 sends to 1, that sends to 2, it would 
be 0 latency, but the opposite would be 128 samples?


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[PD] How latency works in cloned abstraction? (dsp graph order)

2022-01-14 Thread José de Abreu
Hello list, I have a curiosity, let's say that I want to clone an
abstraction that reroutes signals between the cloned instances.

We learn that if a send~ is sorted before the corresponding receive~ we get
no latency, but if the receive~ is sorted before, it reads from the
previous block, so we get by default 64 samples of latency

But how does this order work in the case of cloned abstractions? the
abstractions with a bigger or smaller "index" would be sorted accordingly
(from 0 to n), or is it somewhat random?

so if for example abstraction 0 sends to 1, that sends to 2, it would be 0
latency, but the opposite would be 128 samples?
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