My version of erlang ‘Ring’ problem

I have been reading ‘Erlang Programming‘ , like a good student, I decided to solve the ring exercise:

Here is the problem in my words:

Create a linked list of N Ring where each Ring is an erlang process. When you pass a message to the head Ring , it should propagate the message to the¬† next Ring and that propagates to the next ring and so on. I added additional clause¬† — “Each ring should report its successful message delivery to the previous ring and the previous ring should print successful message delivery from its child rings.

Here is the code:


createRing(N) ->
    Self = self(),
    Head = spawn(fun() -> ring(N, Self) end),   

ring(N, PreviousRing) when N > 0 ->
    Self = self(),
    io:format('~w #ring created PID: [~w] ListenerPid: [~w] ~n',[N, Self, PreviousRing]),
    NextRing = spawn(fun() ->
			     ring(N-1 , Self) end),
    loop(N ,NextRing, PreviousRing);

ring(N, PreviousRing) ->
    io:format('last ring reached! ~w ~n',[N]),
    NullRing = spawn(fun() -> 
				 Any ->
				     io:format('dummy ring received ~w ~n', Any)
    loop(N, NullRing, PreviousRing).

loop(N, NextRing, PreviousRing) ->
	{send_message, Message} ->
	    io:format('ring # ~w with pid: #~w received ~w ~n',[N, self(), Message]),
	    NextRing ! {send_message, Message},
	    PreviousRing ! {ok, self()},
	    loop(N, NextRing, PreviousRing);
	{ok, NextRing} ->
	    io:format('~w with pid# ~w successfully sent message ~n',[N-1, NextRing]),
	    loop(N, NextRing, PreviousRing);
	{quit} ->
	    io:format('quiting ring: ~w with Pid: ~w ~n',[N, self()])

sendMessage(Head, M) when M > 0 ->
    Head ! {send_message, M},
    sendMessage(Head, M - 1);
sendMessage(Head, M) ->
    io:format('sending last message ~n'),
    Head ! {send_message, M}.

I have had my share of multi-threaded programming in Java, thanks to the telephony applications I’v worked on. Just imagine the pain you would have to go through to implement this in Java / C++ !

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