[erlang-questions] Why this code not being infinite recursive?

[I Gusti Ngurah Oka Prinarjaya]

Hi Artie Gold,

Thank you for the additional information about this.
what i do is just for learning purpose. So many syntax variation that i
haven't understand yet.

Thank you


Pada tanggal Sen, 17 Jun 2019 pukul 23.45 Artie Gold <
the.artiegold@REDACTED> menulis:

> But, wait! Infinite (well, unbounded) recursion is a bad thing, right? At
> the very least it will blow up the stack, right?
>
> Well, yes. It would. Except for *tail recursion*. When the recursive step
> is at the end of the function (and there's no lingering state to maintain),
> the compiler can optimize the recursion into a simple loop. As a result you
> get the cleanliness of recursive solution without having to worry about the
> space needed by the calculation being unbounded.
>
> Cheers,
> --ag
>
> On Mon, Jun 17, 2019 at 11:06 AM I Gusti Ngurah Oka Prinarjaya <
> okaprinarjaya@REDACTED> wrote:
>
>> Hi,
>>
>> OMG. Finally..... I know and understand how this code flowing. It takes
>> days for me to understand.
>> It turns out this code below:
>>
>> {ok, AcceptorSocket} = gen_tcp:accept(ListenSocket),
>>
>> is locking / blocking the execution of these lines:
>>
>> 1. spawn(fun() -> acceptor(ListenSocket) end),
>> 2. handle(AcceptorSocket).
>>
>> Then, after we doing a connection to the port via telnet, the locking /
>> blocking become released / unlocked / unblocked, then executing lines: 1
>> and  create new process,
>> lines: 2 listen / waiting the incoming message. Then make a new locking
>> at the new created process for the next telnet connection. Yes this is an
>> infinity recursive.
>> The base case is killing main process that started from start_server/1 . exit(Pid,
>> killed)
>>
>> Thank you :)
>>
>>
>>
>>
>> Pada tanggal Jum, 14 Jun 2019 pukul 11.45 I Gusti Ngurah Oka Prinarjaya <
>> okaprinarjaya@REDACTED> menulis:
>>
>>> Hi,
>>>
>>> I learn gen_tcp from learnyousomeerlangg book. I try to understand flow
>>> of code below.
>>> And i got confused with acceptor/1 . Because acceptor/1 call itself but
>>> have no base case and just execute once. It's not usual. Why?
>>> From my understanding it should be an infinity recursive.
>>>
>>> -module(naive_tcp).
>>> -compile(export_all).
>>>
>>> start_server(Port) ->
>>>   Pid = spawn_link(
>>>     fun() ->
>>>       io:format("Spawned at start_server()~n"),
>>>       {ok, ListenSocket} = gen_tcp:listen(Port, [binary, {active,
>>> false}]),
>>>       spawn(fun() -> acceptor(ListenSocket) end),
>>>       timer:sleep(infinity)
>>>     end
>>>   ),
>>>   {ok, Pid}.
>>>
>>> acceptor(ListenSocket) ->
>>>   io:format("I am acceptor~n"),
>>>   {ok, AcceptorSocket} = gen_tcp:accept(ListenSocket),
>>>   spawn(fun() -> acceptor(ListenSocket) end),
>>>   handle(AcceptorSocket).
>>>
>>> handle(AcceptorSocket) ->
>>>   inet:setopts(AcceptorSocket, [{active, once}]),
>>>   receive
>>>     {tcp, AcceptorSocket, <<"quit", _/binary>>} ->
>>>       gen_tcp:close(AcceptorSocket);
>>>     {tcp, AcceptorSocket, Message} ->
>>>       gen_tcp:send(AcceptorSocket, Message),
>>>       handle(AcceptorSocket)
>>>   end.
>>>
>>> Please enlightenment
>>>
>>> Thank you
>>>
>>>
>>> _______________________________________________
>> erlang-questions mailing list
>> erlang-questions@REDACTED
>> http://erlang.org/mailman/listinfo/erlang-questions
>>
>
>
> --
> Artie Gold, Austin, Texas
> --
> http://makeitsimpler.org
>
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