Oh, I think I realized why that won't work. I think that a program in the State monad, for instance, must execute its computations sequentially. There is no way to "split" a State monad into separate lines of execution from within the monad. I think the answer is that I have to split my computation before my code enters into the State monad. So I need to call runState four times for four cores, or whatever. Something like that.

D


On Fri, Jul 7, 2017 at 1:33 AM, Dennis Raddle <dennis.raddle@gmail.com> wrote:
I'm looking into using rpar and rseq from Control.Parallel.Strategies. The issue is that most of my code executes in a monad stack including StateT and ExceptT. Can I use the Eval monad at the root of the stack safely? 

D

On Thu, Jul 6, 2017 at 10:06 PM, Dennis Raddle <dennis.raddle@gmail.com> wrote:
thanks for the tip, that book is great... I'm reading it now, and it's an easy read, very clear explanations of laziness. I never understood laziness all that well before.

D

On Thu, Jul 6, 2017 at 10:34 AM, Rein Henrichs <rein.henrichs@gmail.com> wrote:
Please take a look at Simon Marlow's free book, Parallel and Concurrent Programming in Haskell (http://chimera.labs.oreilly.com/books/1230000000929). It will teach you a lot about... the things in the title.

On Thu, Jul 6, 2017 at 10:21 AM Claude Heiland-Allen <claude@mathr.co.uk> wrote:
Hi Dennis,

On 06/07/17 17:53, Dennis Raddle wrote:
> I have a program which does backtracking search in a recursive
> function.  I followed the chapter in "Real World Haskell" to parallelize it.
[snip]
> There's no effect from the R.W.H. ideas. Can I get some suggestions as
> to why?

You can get timing and other useful diagnostics by compiling with
-rtsopts and running with +RTS -s, no need to measure CPU time in your
own program.

> force :: [a] -> ()
> force xs = go xs `pseq` ()
>   where go (_:xs) = go xs
>         go [] = 1

This force doesn't do enough, it just walks the spine. Try this, which
forces the elements as well as the shape:

force :: [a] -> ()
force xs = go xs `pseq` ()
  where go (x:xs) = x `pseq` go xs
        go [] = 1

Thanks,


Claude
--
https://mathr.co.uk
_______________________________________________
Haskell-Cafe mailing list
To (un)subscribe, modify options or view archives go to:
http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
Only members subscribed via the mailman list are allowed to post.

_______________________________________________
Haskell-Cafe mailing list
To (un)subscribe, modify options or view archives go to:
http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
Only members subscribed via the mailman list are allowed to post.