[erlang-questions] pattern matching registered name

[James Aimonetti]

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On 05/20/2013 11:47 AM, Edmund Sumbar wrote:
> On Mon, May 20, 2013 at 11:47 AM, James Aimonetti
> <james@REDACTED> wrote:
> 
>> rpc/2 expects the first arg into the function to be the first
>> element in the response tuple. The loop/0 uses self(), which
>> always returns a pid(), not the registered name. So you call rpc
>> with an atom, and receive a tuple with a pid(). Those don't
>> match. Your call to rpc/2 hangs because the selective receive in
>> rpc/2 never matches (and you have no timeout clause).
>> 
> 
> What's confusing is that the atom apparently resolves into a pid()
> in the send expression of kvsx:rpc/2 (adding io:format("reg name =
> ~p~n", [whereis(abc)]) to the function returns the pid()), but not
> in the pattern tuple of the receive expression.
> 
> Anyway, thanks for the replies James and Yogish.
> 
> Ed
> 
> 
> 
> _______________________________________________ erlang-questions
> mailing list erlang-questions@REDACTED 
> http://erlang.org/mailman/listinfo/erlang-questions
> 

You could change rpc/2:

rpc(Name, Q) when is_atom(Name) ->
  rpc(whereis(Name), Q);
rpc(Pid, Q) when is_pid(Pid) ->
  Pid ! {self(), Q},
  receive
    {Pid, Reply} ->
      Reply
  end.

This will crash the caller if Name isn't registered, FYI.

- -- 
James Aimonetti
Distributed Systems Engineer / DJ MC_

2600hz | http://2600hz.com
sip:james@REDACTED
tel: 415.886.7905
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