I use GenServer (restart: :temporary) for a module that manages some state and handles access to it. Its init function and its handle_* functions return the same timeout value. The GenServer-based state-managing processes get started by invoking GenServer.start_link and a :via tuple for the name. I…

I mean this problem is so obvious I agree but the problem exists because we use a Registry for our processes. I think that there is no way to ensure that your process cannot stop in between a fetch on the registry and a call ot the process. I believe you can only mitigate the problem and that pe…

There are multiple solutions to this problem, but it depends on what you want to do. You could simply just call GenServer.call and if you catch a “no process” error then start_link and call again. We need more information on you state : why is it temporary, why at some point you will let it time o…

I am aware the process wouldn’t terminate by itself. That is why I have this handle_info function that receives the :timeout and returns { :stop, :normal, state}. I use GenServer to achieve a sequential access of all change requests over a single instance of a certain data structure (“class”) that …

Thanks for the elaborate reply. I will use try/catch because it’s the lightest of all approaches and I don’t want to take a path that in part replicates the original intent of the whole Erlang multiprocessing paradigm. I just needed to know if I was missing something (I am new to Elixir and Erlang,…

Glad that you found a new home ! Also I forgot to tell, but if the code that calls your registry and then your worker is supervised, and could be restarted issuing the same command, you could also just let the no-process error crash :slight_smile:

Ah yes, the famous “let it fail” paradigm (which I’ve long been in favor of). But in this case, I am not sure whether it will always be the case for the client code, and I like to patch the “holes” as I go, especially the ones I can tell upfront will be difficult to get by accident (except for the …

I feel it is the same : try/cath and retry is the same as crashing the process with noproc and have it be restarted. In both cases your code will fetch the pid again and call your worker (all at once using :via). The latter will repeat more code obviously, but very rarely, and keep the codebase cle…

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GenServer timeout

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DaAnalyst October 18, 2019, 5:30pm 9

Exactly