[erlang-questions] Newbie question. Joe Armstrong book. Ring benchmark.

Russell Brown russell.brown@REDACTED
Thu Sep 25 18:05:02 CEST 2008


On 25 Sep 2008, at 16:06, Edwin Fine wrote:

> Hi Russell,
>
> In general, it's worth searching the erlang archives, which often  
> have the answers to many questions. The searchable archive is at  
> http://www.erlang.org/faq.html.
>
> This post discusses your question. There are quite a few posts in  
> the archives about the ring benchmark.
>
> http://www.erlang.org/pipermail/erlang-questions/2006-March/ 
> 019503.html

Thanks for that. Good resource, should have checked it before, In  
future I will be sure too. Seems my version is like a few others out  
there and the answer is: yeah, ring benchmark is not so concurrent.

Thanks again. Enjoying this.

Russell

>
>
> 2008/9/25 Russell Brown <russell.brown@REDACTED>
>
> Hi,
> I've just done the "ring benchmark" from the end of Chapter 8 in  
> the Joe Armstrong book (Programming Erlang: Software for a  
> Concurrent World).
>
> I think I have a module that does what is asked in the problem  
> "(Write a ring benchmark. Create N processes in a ring. Send a  
> message round the ring M times so that a total of N * M messages  
> get sent. Time how long this takes for different values of N and M.)"
>
> I am certain it is nowhere near the best or most elegant way to do  
> it as I am not very accomplished (but I do like to have a go  
> anyway). What I am wondering is why, when I run the "benchmark"  
> only one of the cores on my laptop's dual core seems to get used.  
> Have I written it wrong (so that there is no concurrency) or is a  
> ring by its nature a sequential thing?
>
> I have attached the code in case it helps answer the above  
> question.  As I say I am a total newb. My reason for posting is  
> just to verify that the implementation works and to find out why it  
> only uses one core.
>
> Also, is there a canonical implementation that beginners like me  
> can have a look at?
>
> Many thanks in advance
>
> Russell
>
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