[erlang-questions] Erlang newbie questions

Gerry Weaver <>
Fri Oct 21 21:04:09 CEST 2011


Hi Garrett,

I write software primarily for banks these days. There are some fairly rigid requirements when it comes to management and monitoring. Applications that start daemon processes from within other daemons processes without dropping a pid file (Unix) for each, are a problem. Windows software virtualization products have the same issue with regard to services. For example, on Unix, when the Erlang VM starts it starts epmd. I didn't see a pid file for either of these. If epmd dies, the VM process will restart it and vice versa. There would need to be pid files created by this case as well. Basically, there would need to be some rework of the start up code. If this were to be reworked, it would make a lot more sense to me to just have these daemons start separately. When it comes to the mutual restarting business, it would be better to have a separate watchdog process for that. This way if
monitoring software is in use, it would fill the monitoring roll and the watchdog process would not need to be started. I also see a major benefit to having the ability to start/control Erlang from within another application, so it would make sense to add that feature, if one was already reworking the code.  I hope this makes sense.


Thanks,
-G



-----Original Message----- 
> From: "Garrett Smith" <> 
> To: "Gerry Weaver" <> 
> Cc:  
> Date: 10/21/11 10:01 
> Subject: Re: [erlang-questions] Erlang newbie questions 
> 
> Hi Gerry,
> 
> On Thu, Oct 20, 2011 at 4:30 PM, Gerry Weaver <> wrote:
> > Hello All,
> >
> > Please disregard the questions in my previous two posts. I have managed to find the answers that I was looking for.
> >
> >
> > Sam made an interesting point with regard to the Unix daemon behavior. This prompted me to dig into to the code a bit more. I'm afraid this would prevent me from using Erlang for any upcoming projects.
> 
> Out of curiosity, with respect to daemon behavior, what requirements
> do you have that can't be met?
> 
> Garrett






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