Many thanks, Chris! I tried this and this works. By the way, Chris, I'm newbie, would you explain - is this right group for Streaming library? And how Streaming is related to Pipes, because author of Streaming (if I understood right) point out this group for questions/discussions, so seems that there is some relation between these 2 libraries. Is it true?
вторник, 30 мая 2017 г., 13:48:16 UTC+3 пользователь Chris Pollard написал: > > Hi Paul, > > You're right: typically when you need to access or modify some state of > type "s" throughout a monadic action, you want your monad to be an instance > of "MonadState s". > > > https://www.stackage.org/haddock/lts-8.15/mtl-2.2.1/Control-Monad-State-Class.html > > Streams are instances of "MonadState s" as long as the underlying monad is > also an instance. Probably the simplest way to do achieve what you need is, > as you mentioned, to change your streams of type "S.Stream (S.Of Int) IO" > to "S.Stream (S.Of Int) (StateT s IO)", where StateT comes from > > > https://www.stackage.org/haddock/lts-8.15/mtl-2.2.1/Control-Monad-State-Strict.html > > With "StateT s IO" as your underlying monad, you can access the current > state with "get" and change it with "modify". > > You will end up with something of type "StateT s IO r" rather than "IO r"; > you can convert this to an IO action using runStateT, evalStateT, or > execStateT, whichever is appropriate. I think you'll find StateT to be > pretty light-weight in terms of performance. > > Cheers, > > Chris > > > > On Monday, May 29, 2017 at 9:10:12 AM UTC+1, aqu...@gmail.com wrote: >> >> Hello, everyone! I'm Haskell beginner and newbie in Streaming library. As >> I understood, this group is dedicated to Pipes, but Streaming library too. >> My simple question is related to Streaming library usage. I'm trying to >> realize next common and popular "pattern": processing of streaming items >> with common state, see Fig, please: >> >> .--state-------+--state'--------+--state''--> >> | | | >> [e0..eN] ==> [e0'...eN'] ==> [e0''..eN''] =====> >> >> >> This "state" will be used for statistics, errors, whatever - through the >> whole workflow. Some of "pipe" nodes will iterate over `eN` items (which >> will be lists/streams too), concatenates results... How this can be >> achieved with Streaming library? I mean each "node" should have access to >> stream items but to "global" state (result of prev. node return?) too. Also >> nodes will do IO actions! >> >> My first attempt was: >> >> ... >> import Streaming >> import qualified Streaming.Prelude as S >> ... >> >> >> -- simulate source of items in the stream >> gen :: S.Stream (S.Of Int) IO [String] >> gen = do >> S.yield 1000 >> S.yield 2000 >> x <- lift getLine -- simulate item got through IO >> return ["a", "b", "c", x] >> >> >> -- simulate some node - processor on the pipe >> proc :: S.Stream (S.Of Int) IO [String] -> S.Stream (S.Of Int) IO [String >> ] >> proc str = do >> e <- str -- WRONG! >> -- S.yield $ e + 8 >> lift $ print "Enter x:" >> x <- lift getLine -- simulate IO communication >> return $ e ++ [" -- " ++ x] -- simulate change of common state via >> result component of ":>" pair >> >> main :: IO () >> main = do >> let >> s <- S.mapM_ print $ S.map show gen >> p <- S.mapM_ print $ proc gen >> putStr "s: " >> print s >> putStr "p: " >> print p >> print "end." >> >> >> But this is wrong sure, because "do e <- str" binds "e" to results of >> previous result, but not to stream's items. And I need to have access to >> both: stream's items and state (which I suppose) can be second component of >> Stream ":>" pair, i.e. result, returning with "return" function. >> >> I tried also to implement "proc" as "a -> m b" function and to apply it >> with "S.mapM", but in this case I'm processing stream's items but without >> previous "return" - state. >> >> Intuitively I'm feeling that solution may be in transformation of result >> monad "m", to be not only IO, but with "StateT", but I don't know how to: >> >> 1. do it >> 2. call it - in drawn workflow >> 3. if I'll yield items wrapped in State monad, what a devil will be >> it, a specially related to performance. >> >> I'll need to iterate over some stream's items too (they are lists), to >> join results and forever to have common state. >> >> Can somebody help with it, please? >> >> === >> Best regards, Paul >> > -- You received this message because you are subscribed to the Google Groups "Haskell Pipes" group. To unsubscribe from this group and stop receiving emails from it, send an email to haskell-pipes+unsubscr...@googlegroups.com. To post to this group, send email to haskell-pipes@googlegroups.com.
