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FreeBSD port build 8.4.3_1 - StackOverflowHook
by Sumit Raja 22 Sep '18

22 Sep '18

Hi, I'm unable to get a consistent build of the FreeBSD 8.4.3 port on a 1.1-RELEASE-p11-amd64. I had success with a one build (in a jail) but am now seeing this problem on a VM: gmake --no-print-directory -f ghc.mk phase=1 phase_1_builds HC [stage 0] ghc/stage1/build/tmp/ghc-stage1 Warning: -rtsopts and -with-rtsopts have no effect with -no-hs-main. Call hs_init_ghc() from your main() function to set these options. /usr/local/lib/ghc-8.4.2/rts/libHSrts.a(RTS.o): In function `StackOverflowHook': /home/ben/bin-dist-8.4.2-FreeBSD/ghc/rts/hooks/StackOverflow.c:16:0: error: multiple definition of `StackOverflowHook' ghc/stage1/build/hschooks.o:hschooks.c:(.text+0x0): first defined here collect2: error: ld returned 1 exit status `gcc7' failed in phase `Linker'. (Exit code: 1) gmake[3]: *** [ghc/ghc.mk:111: ghc/stage1/build/tmp/ghc-stage1] Error 1 Thanks Sumit

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Haskell interpretation of names produces error?
by trent shipley 21 Sep '18

21 Sep '18

data Prop = Const Bool | Var Char | Not Prop | And Prop Prop | If Prop Prop | Or Prop Prop | Yff Prop Prop -- Why do I get errors if "Yff" is replaced with "Iff"? | Xor Prop Prop type Assoc k v = [(k, v)] -- Hutton, Graham. Programming in Haskell (p. 93). Cambridge University Press. Kindle Edition. find :: Eq k => k -> Assoc k v -> v find k t = head [v | (k', v) <- t, k == k'] -- Hutton, Graham. Programming in Haskell (p. 93). Cambridge University Press. Kindle Edition. type Subst = Assoc Char Bool -- Hutton, 2016, Ch 8.6 eval :: Subst -> Prop -> Bool eval _ (Const b) = b eval s (Var x) = find x s eval s (Not p) = not (eval s p) eval s (And p q) = eval s p && eval s q eval s (If p q) = eval s p <= eval s q eval s (Or p q) = eval s p || eval s q eval s (Yff p q) = eval s p == eval s q -- Iff produces error here eval s (Xor p q) = eval s p /= eval s q -- Hutton 2016 Ch 8.6 vars :: Prop -> [Char] vars (Const _) = [] vars (Var x) = [x] vars (Not p) = vars p vars (And p q) = vars p ++ vars q vars (If p q) = vars p ++ vars q vars (Or p q) = vars p ++ vars q vars (Yff p q) = vars p ++ vars q -- Iff produces error here vars (Xor p q) = vars p ++ vars q -- Hutton 2016 Ch 8.6

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Re: [Haskell-beginners] Need better explanation of the 'flipThree' example in LYAH
by Francesco Ariis 19 Sep '18

19 Sep '18

Hello Olumide, On Mon, Sep 17, 2018 at 12:15:58AM +0100, Sola Aina wrote: > flatten( fmap (\a -> ( flatten( fmap ( \b -> return [a,b] ) coin ) ) coin ) > > Is this how to proceed and how am I to simplify this expression? I get a slightly different final expression: flatten (fmap (\a -> flatten (fmap (\b -> return [a,b]) coin)) coin) -- notice the lack of `(` before the second `flatten` This could be rewritten as: flatMap (\a -> flatMap (\b -> return [a,b]) coin) coin -- flatMap :: (a1 -> Prob a2) -> Prob a1 -> Prob a2 -- flatMap f x = flatten (fmap f x) which is a bit easier on the eyes. This expression cannot be simplified any further if we don't bring to the table the definition of `coin`!

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Need better explanation of the 'flipThree' example in LYAH
by Olumide 16 Sep '18

16 Sep '18

Dear List, I'm trying to understand the following example from LYAH import Data.List (all) flipThree :: Prob Bool flipThree = do a <- coin b <- coin c <- loadedCoin return (all (==Tails) [a,b,c]) Where import Data.Ratio newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show and data Coin = Heads | Tails deriving (Show, Eq) coin :: Prob Coin coin = Prob [(Heads,1%2),(Tails,1%2)] loadedCoin :: Prob Coin loadedCoin = Prob [(Heads,1%10),(Tails,9%10)] The result: ghci> getProb flipThree [(False,1 % 40),(False,9 % 40),(False,1 % 40),(False,9 % 40), (False,1 % 40),(False,9 % 40),(False,1 % 40),(True,9 % 40)] See http://learnyouahaskell.com/for-a-few-monads-more#making-monads. My understanding of what's going on here is sketchy at best. One of several explanations that I am considering is that all combination of a, b and c are evaluated in (==Tails) [a,b,c] but I cannot explain how the all function creates 'fuses' the list [f a, f b, f c]. I know that all f xs = and . map f xs (the definition on hackage is a lot more complicated) but, again, I cannot explain how the and function 'fuses' the list [f a, f b, f c]. If I'm on the right track I realize that I'm going to have to study the list the between list comprehensions and the do-notation in order how all the return function create one Prob. Regards, - Olumide

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Hutton Ex 8.9.7
by trent shipley 15 Sep '18

15 Sep '18

I couldn't get close on my own. From: https://github.com/pankajgodbole/hutton/blob/master/exercises.hs {- 7. Complete the following instance declarations: instance Eq a => Eq (Maybe a) where ... instance Eq a => Eq [a] where ... -} -- suggested answer instance Eq a => Eq (Maybe a) where -- Defines the (==) operation. Nothing == Nothing = True Just == Just = True -- why isn't this Just a == Just a ? -- My guess is that a and Just a are different types and can't be == in Haskell _ == _ = False instance Eq a => Eq [a] where -- Defines the (==) operation. [] == [] = True [x] == [y] = x == y (x:xs) == (y:ys) = x==y && xs==ys -- I assume this is implicitly recursive. _ == _ = False

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Hutton Ex 8.9.5 & Ex 8.9.6
by trent shipley 14 Sep '18

14 Sep '18

Pankaj Godbole's code examples for 8.9.6 are sufficiently more complex than mine,that I am worried I missed an obvious bug I'm not experience enough to know about or test for. Correction or reassurances would be appreciated. Regards, Trent {-- 5. Given the type declaration data Expr = Val Int | Add Expr Expr define a higher-order function folde :: (Int -> a) -> (a -> a -> a) -> Expr -> a such that folde f g replaces each Val constructor in an expression by the function f, and each Add constructor by the function g. Hutton, Graham. Programming in Haskell (Kindle Locations 3364-3371). Cambridge University Press. Kindle Edition. --} data Expr = Val Int | Add Expr Expr -- Trent's stuff folde :: (Int -> a) -> (a -> a -> a) -> Expr -> a folde f g (Val x) = f x folde f g (Add left right) = g (folde f g left) (folde f g right) {-- 6. Using folde, define a function eval :: Expr -> Int that evaluates an expression to an integer value, and a function size :: Expr -> Int that calculates the number of values in an expression. Hutton, Graham. Programming in Haskell (Kindle Locations 3372-3374). Cambridge University Press. Kindle Edition. --} -- Trent's stuff eval :: Expr -> Int eval = folde id (+) size :: Expr -> Int size = folde (\_ -> 1) (+) -- hlint suggests \_ as const 1 test0 :: Expr test0 = Val 1 test1 :: Expr test1 = Add (Val 1) (Val 2) test3 :: Expr test3 = Add (Val 1) (Add (Val 2) (Val 3)) -- the below are from: -- Pankaj Godbole -- https://github.com/pankajgodbole/hutton/blob/master/exercises.hs evalE :: Expr -> Int evalE (Val n) = n evalE (Add e1 e2) = folde toEnum (+) (Add e1 e2) -- doesn't a Val have to prefix an int? Then why is f "toEnum"? numVals :: Expr -> Int numVals (Val n) = 1 numVals (Add e1 e2) = numVals e1 + numVals e2

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Why Integral makes a function much slower than Int?
by Jaakko Luttinen 08 Sep '18

08 Sep '18

Hi all, I implemented a simple function to calculate the length of a list. However, I noticed that the speed of the function changes significantly if I make I minor change to the type signature but keep the implementation the same otherwise. I was wondering why this happens. Contents of mylength.hs: module MyLength where -- Here is the first implementation: myLength :: Integral n => [a] -> n myLength xs = run xs 0 where run [] n = n run (_:ys) n = run ys (n+1) -- Then, the other implementation just fixes the return type without -- modifying the actual implementation: myLength' :: [a] -> Int myLength' = myLength When I compile these implementations with: ghc -dynamic -O2 mylength.hs Then, launch ghci: ghci -fobject-code And in ghci: :load mylength.hs :set +s (myLength [1..10000000]) :: Int 10000000 (2.30 secs, 1,612,737,944 bytes) (myLength' [1..10000000]) :: Int 10000000 (0.38 secs, 720,077,536 bytes) So, the first implementation is much worse, although I've fixed the return type to Int too. What's going on? Does this mean one shouldn't use Integral but Int instead most of the time? Thanks for help! Regards, Jaakko

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Alternative ways define trees
by trent shipley 05 Sep '18

05 Sep '18

In which I try a couple of ways to declare alternative trees that can have one or two sub-trees per node, but my attempts don't work. How do I get them to work? Is it preferable to require two trees per node? Regards, Trent Shipley {-- 3. Consider the following type of binary trees: data Tree a = Leaf a | Node (Tree a) (Tree a) Let us say that such a tree is balanced if the number of leaves in the left and right subtree of every node differs by at most one, with leaves themselves being trivially balanced. Define a function balanced :: Tree a -> Bool that decides if a binary tree is balanced or not. Hint: first define a function that returns the number of leaves in a tree. Hutton, Graham. Programming in Haskell (Kindle Locations 3355-3361). Cambridge University Press. Kindle Edition. --} data Tree a = Leaf a | Node (Tree a) (Tree a) -- From Hutton. -- You can't have an empty tree of a node with no branches. A degenerate tree consists of one leaf, when my intuition says a leaf should be required to have a node. A node must have two sub-trees, it can't have one sub-tree. I have mixed feelings about not being able to have a node with one sub-tree. data Tree' a = Leaf' a | Node' (Tree' a) Maybe (Tree' a) -- First failed experiment. -- can this be made with the "data" word, if not what are the options? This is my preferred constructor. A node that can contain two trees, or one tree and "Nothing". {------------------------- ex8_3experiments.hs:20:46: error: • Expecting one more argument to ‘Maybe’ Expected a type, but ‘Maybe’ has kind ‘* -> *’ • In the type ‘Maybe’ In the definition of data constructor ‘Node'’ In the data declaration for ‘Tree'’ Failed, modules loaded: none. ---------------------------} data Tree'' a = Leaf'' a | Node'' (Tree'' a) (Tree'' a) | Node'' (Tree'' a) -- second failed experiment. -- This is an inferior option for this, since your tree walking functions have to worry about a non-existent right path. -- How would I create this? {-------------------------- ex8_3experiments.hs:21:59: error: Multiple declarations of ‘Node''’ Declared at: ex8_3experiments.hs:21:28 ex8_3experiments.hs:21:59 Failed, modules loaded: none. ---------------------------} tEven :: Tree Int tEven = Node (Node (Leaf 1) (Leaf 2)) (Node (Leaf 3) (Leaf 4)) -- Test :: Tree Int -- tTest = Node (Node (Leaf 1) (Leaf 2)) (Node (Leaf 3)) {--------------- To make a "canonical" balanced odd tree you have an odd number of two leaf sub-trees, and one three leaf sub-tree at the end. n / \ / \ n n / \ / \ 1 2 3 n / \ 4 5 ----------------} tOdd :: Tree Int tOdd = Node (Node (Leaf 1) (Leaf 2)) (Node (Leaf 3) (Node (Leaf 4) (Leaf 5))) tUnbal :: Tree Int tUnbal = Node (Node (Leaf 1) (Node (Leaf 2) (Node (Leaf 3) (Leaf 4)))) (Node (Leaf 5) (Leaf 6)) leaves :: Tree a -> Int leaves (Leaf _) = 1 leaves (Node l r) = leaves l + leaves r -- From Hutton's solutions. balanced :: Tree a -> Bool balanced (Leaf _) = True balanced (Node l r) = abs(leaves l - leaves r) <= 1 && balanced l && balanced r -- From Hutton's solutions.

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Hutton 2016 ex8.3a
by trent shipley 04 Sep '18

04 Sep '18

Given data Tree a = Leaf a | Node (Tree a) (Tree a) Write a leaf counter. Hutton suggests: leaves :: Tree a -> Int leaves (Leaf _) = 1 leaves (Node l r) = leaves l + leaves r I tried: leavesTrent :: Tree a -> Int leavesTrent = leaves' 0 where leaves' n (Leaf a) = n + 1 leaves' n (Node l r) = (leaves' n l), (leaves' n r) The idea is: If it is a leaf, add one to the accumulator. (Following Hutton's explanation of how sum works if defined with foldl.) If it is a tree, proceed down the left subtree recursively, until you get to a leaf, then roll up to the right subtree. The problem (among the problems) is that I don't know how to tell the compiler to do all lefts, before backing up to go right. I only know how to do that using a real operator like "+" or foo (l, r). Is that kind of no-op recursive branching possible? Trent.

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Attoparsec parser and maybe type
by PICCA Frederic-Emmanuel 29 Aug '18

29 Aug '18

Hello, I have a type compose of 5 doubles data MyData = MyData Double Double Double Double Double now I have a file to parse whcih contain this | 1.0 2.0 3.0 4.0 5.0 | or | ---------no result ------------- | So I would like to create a parser which return a Maybe Mydata | 1.0 2.0 3.0 4.0 5.0 | -> Just MyData | ---------no result ------------- | -> Nothing How should I proceed I can write really trivially both parser but I do not know how to combine all this MyDataP1 = MyData <$> scientific <*> scientific NothingP = string "--------------- no result ----------------" Thanks for your help Frederic

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