April 2016 - Beginners

Trying to install "Craft3e" from Hackage using stack
by Albin Otterhäll 16 Apr '16

16 Apr '16

I'm currently reading "Haskell: the Craft of Functional Programming", by Simon Thompson and I'm trying to download and install the package books code. But when I run "stack build Craft3e" I get the following error message: > Run from outside a project, using implicit global project config > Using resolver: lts-5.12 from implicit global project's config file: > /home/dumbl3d0re/.stack/global-project/stack.yaml > While constructing the BuildPlan the following exceptions were encountered: > > -- Failure when adding dependencies: > HUnit: needed (==1.2.*), 1.3.1.1 found (latest applicable is 1.2.5.2) > mtl: needed (>=1.1 && <2.2), 2.2.1 found (latest applicable is 2.1.3.1) > needed for package Craft3e-0.1.0.10 How should I go about to install "Craft3e"? Thanks in advance! Regards, Albin

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Unicode characters in function names: some don't work?
by Silent Leaf 14 Apr '16

14 Apr '16

Hi! say I wanna use "—" as new infix operator: it's the big dash used a bit like parenthesis, especially at the end of sentences —like this. in ghci directly: Prelude> let (—) a b = a + b No problem, is accepted and usable. Same in files. Now I try using (»), a French (amongst others) punctuation sign, typically replaces the quote-ends, « like this ». Doesn't work: <interactive>:2:6: lexical error at character '\187' I thought Haskell was Unicode-friendly? Why some symbols but not others? :'(

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function application
by mike h 12 Apr '16

12 Apr '16

Hi, I’m looking at mc :: (Integral a) => a -> a mc x | x > 100 = x - 10 | otherwise = mc ( mc ( x + 11 ) ) the mc ( mc ( x + 11 ) ) can also be written as mc . mc $ x + 11 and I expected it could also be written as mc . mc ( x + 11 ) but the compiler error starts off with Couldn't match expected type ‘a’ with actual type ‘a0 -> c0’ so that is telling me, isn’t it (?) , that using parens is making the argument to the second mc into a function ‘a0 -> c0’ So is mc . mc $ x + 11 the only correct way to write this particular function in ‘.’ style ? Many Thanks Mike

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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?

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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. :) >

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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

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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

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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

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