[Haskell-beginners] Count with Writer asynchronously
Baa
aquagnu at gmail.com
Tue Jul 25 11:00:10 UTC 2017
Yes, it's true. I see now. Thank you again for that good
explanation!
On Tue, 25 Jul 2017 13:41:35 +0300
Michael Snoyman <michael at snoyman.com> wrote:
> You're asking very good questions, about a topic I considered
> bringing up in my previous response but didn't. I guess I will now :)
>
> The important thing to realize about monad transformers (and monads in
> general) is that _they're nothing special_. Under the surface,
> WriterT is simply:
>
> * Modifying your functions to have an extra value returned
> * Providing convenience functions like `tell` and `liftIO` to work
> with that extra value
>
> I would strongly advise getting used to writing code in both the
> transformer and non-transformer style to convince yourself that they
> are the same thing. To make this concrete, look at the WriterT
> definition:
>
> newtype WriterT w m a = WriterT { runWriterT :: m (a, w) }
>
> If you take away the newtype wrapping business, you can see that it's
> just saying `WriterT w m a = m (a, w)`. And therefore, if we go back
> to your types, we can replace:
>
> WriterT (Sum a) IO ()
>
> with
>
> IO ((), Sum a)
>
> In fact, that's exactly what `runWriterT` is doing. I would recommend
> trying to rewrite your code to not even bother with the `WriterT` at
> all and see if you can get the same output.
>
> So your intuition is correct: there's no output parameter like in C.
> You could simulate that in Haskell by passing in a mutable variable
> (like an IORef), but that's not idiomatic. Passing back pairs of
> values is common in Haskell.
>
> Transformers can be useful for avoiding a lot of boilerplate code.
> They can also introduce a lot of complexity in some cases. Figuring
> out when is the right time to use them and when not is subjective and
> nuanced. Knowing the skill is great; being able to avoid using the
> skill is also important :)
>
> On Tue, Jul 25, 2017 at 1:29 PM, Baa <aquagnu at gmail.com> wrote:
>
> > Hello, Michael! This answers to my question completely, thank you so
> > much!
> >
> > But looking at this code, I thought: how is it right to wrap/unwrap
> > write monad? In languages like C or Python we can pass input/output
> > argument (`int*` in C or `[0]` in Python) and use it as some
> > accumulator. But here writer monad is not using as accumulator,
> > accumulating (summation) happens in `mconcat`, right? It's using
> > only as output value, i.e. place to "yield" result. I mean `w` is
> > 0, each call of `runIt` sets there 1, after all calls we calculate
> > sum of this 1's. And instead of `censor (+1) w` I can do `tell 1`
> > only.
> >
> > It means that `runIt` can return not `IO ()` but `IO Int` and
> > results of all `runIt`'s asynchnronously gotten values can be
> > accumulated with `mconcat` without using of writer monad. Am I
> > right, writer monad here is not accumulator but only output value
> > (like output arguments in C/C++/IDL/etc)? How is this a typical
> > solution in Haskell - to use writer monad with wrap/unwrap multiple
> > times, only to save output value?
> >
> >
> >
> > On Tue, 25 Jul 2017 12:31:56 +0300
> > Michael Snoyman <michael at snoyman.com> wrote:
> >
> > > Firstly, a direct answer to your question: use mconcat.
> > >
> > > main :: IO ()
> > > main = do
> > > let l = [1,2,3,4]
> > > w = writer ((), 0) :: WriterT (Sum Int) IO ()
> > > z <- sequence
> > > (map (A.async . runWriterT . runIt w) l)
> > > >>= mapM A.wait
> > > print $ snd $ mconcat z
> > >
> > > Under the surface, WriterT is using mappend to combine the `Sum`
> > > values anyway, so it's natural is `mconcat` (the version of
> > > mappend that applies to list) to get the same result. Now some
> > > possible improvements.
> > >
> > > You're not actually using the return value from the `runIt` call,
> > > just the writer value. There's a function called `execWriter` for
> > > this:
> > >
> > > z <- sequence
> > > (map (A.async . execWriterT . runIt w) l)
> > > >>= mapM A.wait
> > > print $ mconcat z
> > >
> > > Next, the combination of map and sequence can be written as
> > > traverse:
> > >
> > > z <- traverse (A.async . execWriterT . runIt w) l
> > > >>= mapM A.wait
> > >
> > > But the async library is cool enough that it provides a function
> > > called mapConcurrently that deals with the async/wait dance for
> > > you:
> > >
> > > main :: IO ()
> > > main = do
> > > let l = [1,2,3,4]
> > > w = writer ((), 0) :: WriterT (Sum Int) IO ()
> > > z <- A.mapConcurrently (execWriterT . runIt w) l
> > > print $ mconcat z
> > >
> > > One final note: usage of `print` like this in a concurrent context
> > > can run into interleaved output if you have the wrong buffer mode
> > > turned out, leading to output like this:
> > >
> > > 2
> > > 3
> > > 41
> > >
> > > This is especially common when using runghc or ghci. You can
> > > either change the buffering mode or use a different output
> > > function like sayShow (from the say package, which I wrote):
> > >
> > > module Main where
> > >
> > > import qualified Control.Concurrent.Async as A
> > > import Control.Monad.Trans.Writer
> > > import Data.Monoid
> > > import Say
> > >
> > > runIt :: (Show a, Num a)
> > > => WriterT (Sum a) IO ()
> > > -> a
> > > -> WriterT (Sum a) IO ()
> > > runIt w x = do
> > > censor (+1) w -- emulates conditional count of something
> > > sayShow x
> > >
> > > main :: IO ()
> > > main = do
> > > let l = [1,2,3,4]
> > > w = writer ((), 0) :: WriterT (Sum Int) IO ()
> > > z <- A.mapConcurrently (execWriterT . runIt w) l
> > > sayShow $ mconcat z
> > >
> > >
> > > On Tue, Jul 25, 2017 at 11:36 AM, Baa <aquagnu at gmail.com> wrote:
> > >
> > > > Hello, Dear List!
> > > >
> > > > There is package `async`
> > > > (https://hackage.haskell.org/package/async-2.1.1.1/docs/
> > > > Control-Concurrent-Async.html).
> > > >
> > > > Before, I had:
> > > >
> > > > import qualified Control.Concurent.Async as A
> > > > ...
> > > > runIt :: ... -> IO ()
> > > > ...
> > > > sequence [A.async $ runIt ...] >>= mapM_ A.wait
> > > >
> > > > But now I want to count something inside `runIt`. I will use
> > > > `Writer` monad for it (sure, it can be `State` also, not in
> > > > principle to me). To do it synchronously, I done:
> > > >
> > > > module Main where
> > > >
> > > > import Control.Monad.Trans.Writer
> > > > import Control.Monad.IO.Class
> > > > import Data.Monoid
> > > >
> > > > runIt :: (Show a, Num a) => WriterT (Sum a) IO () -> a ->
> > > > WriterT (Sum a) IO ()
> > > > runIt w x = do
> > > > censor (+1) w -- emulates conditional count of something
> > > > liftIO $ print x
> > > >
> > > > main = do
> > > > let l = [1,2,3,4]
> > > > w = writer ((), 0) :: WriterT (Sum Int) IO ()
> > > > z <- runWriterT $ sequence [runIt w i | i <- l]
> > > > print $ snd z
> > > >
> > > > but now my `runIt` changes it's signature:
> > > >
> > > > runIt :: Num a => WriterT (Sum a) IO () -> ... -> WriterT
> > > > (Sum a) IO () ...
> > > > sequence [A.async $ runIt ...] >>= mapM_ A.wait
> > > > ^^^^^^^^^^^
> > > > ` ERROR is here!
> > > >
> > > > I get the error because `async`::IO () -> IO (A.Async ()) but
> > > > I'm trying to pass it `WriterT (Sum a) IO ()`!
> > > >
> > > > To fix it I added `runWriterT` there:
> > > >
> > > > res <- sequence [A.async $ runWriterT (runIt ...) ...] >>=
> > > > mapM A.wait
> > > >
> > > > but now I will get list of counters, not one (res::[((), Sum
> > > > Int)])!
> > > >
> > > > How to solve this problem: to run several actions asyncronously
> > > > and to count something
> > > > inside the action with `Writer` monad?
> > > >
> > > >
> > > > ===
> > > > Best regards, Paul
> > > > _______________________________________________
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> > > > Beginners at haskell.org
> > > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> > > >
> >
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