- Beginners - Haskell.org

Re: [Haskell-beginners] Haskell triangular loop (correct use of (++))
by Silent Leaf 12 Apr '16

12 Apr '16

I'm sorry to hear, no implicit memoization (but then is there an explicit approach?). in a pure language, this seems hardly logical, considering the functional "to one output always the same result" and the language's propensity to use recursion that uses then previous values already calculated. Really hope for an explicit memoization! and i don't mean a manual one ^^ if that is even possible? Anyway, i just don't get your function f. you did get that in mine, xs was the value of my comprehension list, aka [.s1..n] ++ [0] >From there, if I'm not wrong, your function creates a list of all truncated lists, I supposed to be used with a call for the nth element? True, it memorizes all needed things, and in theory only "drops" once per element of the list. As for incorporating it, i'm thinking, local variable? ^^ the function itself could be outside the comprehension list, in a let or where, or completely out of everything, I don't really see the problem. then it's just a matter of it being called onto xs inside the comprehension list, like that: result = [(x,y) | let xs = [1,2,3,0], let yss = f xs, (x,i) <- zip xs [1,2..], y <- yss !! i, x /= y] The remaining possible issue is the call to !!... dunno if that's costy or not. The best way to go through the list I suppose would be by recursion... oh wait, I'm writing as I'm thinking, and I'm thinking: result = [(x,y) | let xs = [1,2,3,0], let yss = f xs, x <- xs, ys <- yss, y <- ys, x /= y] after all, why not use the invisible internal recursion? What do you think? As for the number-instead-of-number-list, the crucial point i mentioned was to determine the maximum number of digit (biggest power of ten reached), and fill up the holes of the smaller numbers with zeroes, so your examples would be: [1,2,3] = 123 --maximum size of a number = 1, no need to fill up [12,3] = 1203 --maximum size of a number = 2 digits, thus the hole beside 3 gets filled with a zero, just like on good old digital watches ^^ Do you think extraction of clusters of digits from numbers would be advantageous, efficiently speaking?

3 3

Re: [Haskell-beginners] does this function exist already? (a -> b -> c) -> (d -> b) -> (( a -> d -> c ))
by Silent Leaf 08 Apr '16

08 Apr '16

Thanks! I'll definitely look into it. :) >

1 0

does this function exist already? (a -> b -> c) -> (d -> b) -> (( a -> d -> c ))
by Silent Leaf 08 Apr '16

08 Apr '16

[Note, total newbie to mailing lists in general, and beginner in haskell as one might have surmised ^^ hopefully I'm doing the right thing here with this mail...] All, or virtually all in the title, but I'll develop. I wrote some useful function (which i rather awkwardly called "point2", for "(.) operator applied at the level of the second argument of the first function"). Here is its def: point2 :: (a -> b -> c) -> (d -> b) -> (( a -> d -> c )) -- the double parentheses, totally optional and syntactically pointless, are just there to imply that the function is seen by we users as 'outputting' another function, even though said outputted function can be applied right away, of course, outputting then possibly c, or (d -> c) if partial application. -- I named the variables in the pattern like their respective types, for 'clarity' point2 f g a d = f a (g d) -- i originally wrote it in point free style using flip, from Data.Function (well, my own rebuilt wheel actually); so, alternate definition using flip == (\f y x -> f x y) point2 f g = flip (flip f . g) -- proof by decomposition: (or is it "composition", as we don't actually decompose?) -- f :: a -> b -> c -- flip f :: b -> a -> c -- flip f . g :: d -> a -> c -- flip (flip f . g) == f `point2` g :: a -> d -> c anyways, it's pretty useful if you wanna combine f and g but g is meant to output the *second* argument of f, not the first, that is: (f a) . g == (f `point2` g) a in the first expression, "a" must necessarily be known and given to combine f and g in point-free style, whereas in the second one, "a" can be omitted, hence then we could write "foo = f `point2` g", which is way closer to the point-free style, way simpler to understand too in my opinion once you got the picture. If you got all I said above, my question is then to know if this point2 function already exists officially, coz I don't really wanna reinvent the wheel, plus I wonder how they called it ^^ I'm not really satisfied of "point2" as variable name. I'd love (.2) but it's not compatible with Haskell. ^^ Also, same question for the following function (does it already exists?), again a sibling of (.), here, the purpose being to write h(a, b) = f (g(a, b)) in point-free style: after2 :: (c -> d) -> (a -> b -> c) -> (( a -> b -> d )) after2 f g a b = f (g a b) -- its name implies an infix use, for example: h = f `after2` g basically, if you know about Data.Function(on), it's a bit (one of) its opposite: `on` applies g to both arguments of f independently, before giving both results to the 2-ary function f, ie (f `on` g) a b == f (g a) (g b) I'm not entirely sure, but I think we could write: f `after2` g == curry (f . (uncurry g)) -- since: uncurry g :: (a,b) -> c f . uncurry g :: (a,b) -> d curry (f . uncurry g) :: a -> b -> d un/curry functions' defs, if needed: curry :: ((a,b) -> c) -> a -> b -> c curry f a b = f (a,b) -- curry f :: a -> b -> c when f :: (a,b) -> c uncurry :: (a -> b -> c) -> (a,b) -> c uncurry g (a,b) = g a b -- uncurry g :: (a,b) -> c when g :: a -> b -> c

2 3

Queues in Haskell
by rohan sumant 06 Apr '16

06 Apr '16

Hello, According to http://rafal.io/posts/haskell-queues.html (++) cant be used to implement queues in Haskell. The reason being that a push operation takes linear time which is indeed very valid given that (++) operates in linear time. However, since Haskell is lazy shouldn't (++) be implemented only when the need occurs? In head ([1..] ++ [10]) I sincerely doubt the the [10] is concatenated before evaluating the head of the list. Please note that this question is focused upon the internals of Haskell. I am aware that there are other approaches to implementing queues in Haskell. Rohan Sumant

2 2

How to write faster ByteString/Conduit code
by John Ky 04 Apr '16

04 Apr '16

Hello Haskellers, I’ve been trying to squeeze as much performance out of my code as possible and I’ve come to a point where can’t figure out what more I can do. Here is some example code: blankEscapedChars :: MonadThrow m => Conduit BS.ByteString m BS.ByteString blankEscapedChars = blankEscapedChars' "" blankEscapedChars' :: MonadThrow m => BS.ByteString -> Conduit BS.ByteString m BS.ByteString blankEscapedChars' rs = do mbs <- await case mbs of Just bs -> do let cs = if BS.length rs /= 0 then BS.concat [rs, bs] else bs let ds = fst (unfoldrN (BS.length cs) unescapeByteString (False, cs)) yield ds blankEscapedChars' (BS.drop (BS.length ds) cs) Nothing -> when (BS.length rs > 0) (yield rs) where unescapeByteString :: (Bool, ByteString) -> Maybe (Word8, (Bool, ByteString)) unescapeByteString (wasEscaped, bs) = case BS.uncons bs of Just (_, cs) | wasEscaped -> Just (wUnderscore, (False, cs)) Just (c, cs) | c /= wBackslash -> Just (c, (False, cs)) Just (c, cs) -> Just (c, (True, cs)) Nothing -> Nothing The above function blankEscapedChars will go find all \ characters and convert the following character to a _. For a 1 MB in memory JSON ByteString, it benches at about 6.6 ms In all my code the basic strategy is the same. await for the next byte string, then use and unfoldrN to produce a new ByteString for yielding. Anyone know of a way to go faster? Cheers, -John

1 2

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?

1 0

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

2 3

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

4 6

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…

2 2

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

3 4