March 2016 - Beginners

Multiline code in GHCi
by Shishir Srivastava 29 Aug '17

29 Aug '17

Hi, I am trying to run this basic multiline code in GHCi. I am not sure if this is even possible without writing it in a file and compiling it in a traditional way. ------- Prelude> :set +m Prelude> let main1 = do Prelude| a <- readLn Prelude| return a Prelude| Prelude> Prelude> main1 <interactive>:75:1: No instance for (Show (IO b0)) arising from a use of `print' In the first argument of `print', namely `it' In a stmt of an interactive GHCi command: print it ------- Could someone please confirm if the function can be defined and called like this in GHCi or point to where am going wrong. Thanks, Shishir Srivastava

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Type constraints of type family arguments
by Dmitriy Matrosov 30 Mar '16

30 Mar '16

> {-# LANGUAGE DataKinds, KindSignatures, PolyKinds, TypeFamilies, GADTs, StandaloneDeriving #-} Hi. I've tried to implement type-level list myself (as an exercise, not looking into existing libraries): Here are two variants of singletons: > data SList1 :: [*] -> * where > SNil1 :: SList1 '[] > SCons1 :: a -> SList1 t -> SList1 ('(:) a t) > > data SList2 (a :: *) (t :: [*]) :: * where > SNil2 :: SList2 a '[] > SCons2 :: a -> SList2 a t -> SList2 a ('(:) a t) > deriving instance Show a => Show (SList2 a b) and `map` for type lists: > type family MapF (b :: r) (t :: [k]) :: [r] where > MapF b '[] = '[] > MapF b ('(:) a t) = ('(:) b (MapF b t)) Then i can write `map` on second singleton variant: > mapF2 :: (a -> b) -> SList2 a t -> SList2 b (MapF b t) > mapF2 f SNil2 = SNil2 > mapF2 f (SCons2 x xs) = SCons2 (f x) (mapF2 f xs) but it does not type-check on first: mapF1 :: (a -> b) -> SList1 t -> SList1 (MapF b t) mapF1 f SNil1 = SNil1 mapF1 f (SCons1 x xs) = SCons1 (f x) (mapF1 f xs) because (as i understand) ghc can't deduce from type-signature, that list element type `a` should match type `a` of first map function's argument: Could not deduce (a1 ~ a) from the context (t ~ (a1 : t1)) bound by a pattern with constructor SCons1 :: forall a (t :: [*]). a -> SList1 t -> SList1 (a : t), in an equation for ‘mapF1’ at 1.lhs:37:12-22 ‘a1’ is a rigid type variable bound by a pattern with constructor SCons1 :: forall a (t :: [*]). a -> SList1 t -> SList1 (a : t), in an equation for ‘mapF1’ at 1.lhs:37:12 ‘a’ is a rigid type variable bound by the type signature for mapF1 :: (a -> b) -> SList1 t -> SList1 (MapF b t) at 1.lhs:35:12 Relevant bindings include x :: a1 (bound at 1.lhs:37:19) f :: a -> b (bound at 1.lhs:37:9) mapF1 :: (a -> b) -> SList1 t -> SList1 (MapF b t) (bound at 1.lhs:36:3) In the first argument of ‘f’, namely ‘x’ In the first argument of ‘SCons1’, namely ‘(f x)’ Then i try to bind these types using different type family definition: > type family MapF' (a :: k) (b :: r) (t :: [k]) :: [r] where > MapF' a b '[] = '[] > MapF' a b ('(:) a t) = ('(:) b (MapF' a b t)) but ghc still can't deduce type equality: mapF1' :: (a -> b) -> SList1 t -> SList1 (MapF' a b t) mapF1' f SNil1 = SNil1 mapF1' f (SCons1 x xs) = SCons1 (f x) (mapF1' f xs) Could not deduce (a1 ~ a) from the context (t ~ (a1 : t1)) bound by a pattern with constructor SCons1 :: forall a (t :: [*]). a -> SList1 t -> SList1 (a : t), in an equation for ‘mapF1'’ at 1.lhs:72:13-23 ‘a1’ is a rigid type variable bound by a pattern with constructor SCons1 :: forall a (t :: [*]). a -> SList1 t -> SList1 (a : t), in an equation for ‘mapF1'’ at 1.lhs:72:13 ‘a’ is a rigid type variable bound by the type signature for mapF1' :: (a -> b) -> SList1 t -> SList1 (MapF' a b t) at 1.lhs:70:13 Relevant bindings include x :: a1 (bound at 1.lhs:72:20) f :: a -> b (bound at 1.lhs:72:10) mapF1' :: (a -> b) -> SList1 t -> SList1 (MapF' a b t) (bound at 1.lhs:71:3) In the first argument of ‘f’, namely ‘x’ In the first argument of ‘SCons1’, namely ‘(f x)’ And i noticed that `MapF'` type family does not enforce constraint on list element type: *Main> :t undefined :: SList1 (MapF' Int Char ('(:) Int '[])) undefined :: SList1 (MapF' Int Char ('(:) Int '[])) :: SList1 '[Char] but this should (shouldn't it?) be an error *Main> :t undefined :: SList1 (MapF' Float Char ('(:) Int '[])) undefined :: SList1 (MapF' Float Char ('(:) Int '[])) :: SList1 (MapF' Float Char '[Int]) only constraint on its kind: *Main> :t undefined :: SList1 (MapF' ('Just Int) Char ('(:) Int '[])) The third argument of ‘MapF'’ should have kind ‘[Maybe *]’, but ‘(:) Int '[]’ has kind ‘[*]’ In an expression type signature: SList1 (MapF' (Just Int) Char ((:) Int '[])) In the expression: undefined :: SList1 (MapF' (Just Int) Char ((:) Int '[])) Is there a way to restrict list element type to match the type of first MapF' type family's argument? And is there a way to write `map` on SList1?

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bracket or how to release ressources.
by PICCA Frederic-Emmanuel 29 Mar '16

29 Mar '16

Hello, I try to write a sort of withxxx method whcih release a file withH5File :: FilePath -> (HId_t -> IO r) -> IO r withH5File name = bracket (withCString name acquire) h5f_close where acquire file = h5f_open file h5f_ACC_RDONLY h5p_DEFAULT In my case h5f_open return a HId_t value (HId_t Int) If something went wrong the returned value is negativ. In that case I do not need to h5f_close the file but instead return an error My question is how should I write the bracket part when the finaliser (h5f_close) depends on the result of the acquire part. thanks for your help Frederic

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let x = x in x (GHC.Prim)
by John Ky 23 Mar '16

23 Mar '16

Hello Haskellers, I'm trying to write a faster popCount function for x86 systems. I tried cloning the ghc-prim package and repurposing it for my own needs, but it isn't working as hoped. In particular, popCnt64# was implemented in GHC.Prim as: popCnt64# = let x = x in x Which shouldn't terminate. Yet when I call it, it magically finds the C implementation in hs_popcnt64 and returns the correct value. My cloned project doesn't behave that way. Instead it doesn't terminate as I would expect. Anyone know what's happening here, if there is a way to make this work or tell me if I'm going about this completely the wrong way? Cheers, -John

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the flag -N2 requires the program to be built with -threaded
by Jung Kim 23 Mar '16

23 Mar '16

I am currently reading an article[1] where the author uses Haskell to explain her idea. The problem is I don't understand Haskell and am not familiar with Haskell concept like MVar, IVar, etc. (I can get the code compile, and have several years programming experiences.) When testing the code, it throws the error messages "the flag -N2 requires the program to be built with -threaded"; however I can't find thread related flag with ghc command by `ghc --help`. How can I fix this problem? Environment I use: Debian stretch/sid, kernel 4.0.0-2-rt-686-pae, GHC version 7.10.3. The code is exactly the same as described in the section 'Nondeterminism with MVars' of article; and the makefile content is all: ghc -O mvar.hs ghc -o mvar mvar.o run: ./execute # while true; do ./mvar +RTS -N2; done clean: rm mvar Thanks [1]. http://composition.al/blog/2013/09/22/some-example-mvar-ivar-and-lvar-progr…

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How to call popCnt64#?
by John Ky 22 Mar '16

22 Mar '16

Hello Haskellers, Does anyone know how to call popCnt64# from the GHC.Prim module? This was my failed attempt: λ> popCnt64# 1 <interactive>:14:11: Couldn't match kind ‘*’ with ‘#’ When matching types a0 :: * Word# :: # Expected type: Integer -> Word# Actual type: Integer -> a0 In the first argument of ‘popCnt64#’, namely ‘1’ In the expression: popCnt64# 1 In an equation for ‘it’: it = popCnt64# 1 -John

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pipe and IO
by PICCA Frederic-Emmanuel 22 Mar '16

22 Mar '16

Hello, Still playing with the Pipe module here the code I try to write getDataFrameHkl3D' :: DataFrameHkl3DH5SixsUhv -> Int -> IO DataFrameHkl3D getDataFrameHkl3D' d i = do positions <- get_position (h5mu d) i return $ DataFrameHkl3D { df_n = i , df_image = positions } getDataFrameHkl3D :: DataFrameHkl3DH5SixsUhv -> Producer DataFrameHkl3D IO () getDataFrameHkl3D d = do n <- lift $ hkl_h5_len (h5mu d) forM_ [0..n] $ \i -> yield $ getDataFrameHkl3D' d i but when I compile it I get this error message src/hkl3d.hs:274:3: Couldn't match type `IO DataFrameHkl3D' with `DataFrameHkl3D' Expected type: Proxy X () () DataFrameHkl3D IO () Actual type: Proxy X () () (IO DataFrameHkl3D) IO () In a stmt of a 'do' block: forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i In the expression: do { n <- lift $ hkl_h5_len (h5mu d); forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i } In an equation for `getDataFrameHkl3D': getDataFrameHkl3D d = do { n <- lift $ hkl_h5_len (h5mu d); forM_ [0 .. n] $ \ i -> yield $ getDataFrameHkl3D' d i } I do not know how to fix my code... Frederic

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haskell + python numpy
by PICCA Frederic-Emmanuel 22 Mar '16

22 Mar '16

Hello, I would like to use some python code in order to do my computation with haskell. The idea is to use python and all it's numrical stack in some places, until I find the time to replace the python code by some haskell code. what I will do is read the data from hdf5 with bindings-hdf5 in haskell obtain a Storable for all my data images then I would like to create a python object which will use these data in order to generate in return another array. get back this array and put it into another hdf5 file. Indeed I want to do this with the pipes library. I would like to know if someone have some code examples for this kind of use case. interfacing haskell with python numpy object back and forth. thanks for your help Frederic

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Do IO actions *have* to be glued by the do syntax?
by Olumide 22 Mar '16

22 Mar '16

Hello List, Do IO actions *have* to be glued by the do syntax? Many, if not all, the examples that I've come across in LYH seem to suggest so. And if so, why? BTW, if possible I'd appreciate an explanation that does not resort to monads. I haven't studied them yet and I'm sure I'd struggle to understand any explanation that resorts to monads. Thanks, - Olumide

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Need help groking the statement: expression1 in term:expression2
by Olumide 22 Mar '16

22 Mar '16

Hello List, I'd appreciate help understanding the second line of following block of code (from LYH, first line added for completeness), http://learnyouahaskell.com/input-and-output#randomness randoms' :: (RandomGen g, Random a) => g -> [a] randoms' gen = let (value, newGen) = random gen in value:randoms' newGen The part I'm really struggling with is random gen in value:randoms' newGen Thanks, - Olumide

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