[Haskell-cafe] parallel Haskell

[Dennis Raddle]

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 at 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 at 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 at gmail.com>
>> wrote:
>>
>>> Please take a look at Simon Marlow's free book, *Parallel and
>>> Concurrent Programming in Haskell* (http://chimera.labs.o
>>> reilly.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 at 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
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>>
>
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