April 2015 - Beginners

Using Show
by Shishir Srivastava 16 Apr '15

16 Apr '15

hi, I am trying to naively print an applicative functor using 'show' like this - show (pure (+) <*> [1, 2, 3]) I know there is something fundamentally wrong with that expression but not sure what. Can anyone please point it out. Thanks, Shishir

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Parse error
by Shishir Srivastava 16 Apr '15

16 Apr '15

Hi, I am trying to run the following from the GHCi command prompt which results in parse error. -------------- Prelude> :{ Prelude| foo :: Maybe String Prelude| foo = do Prelude| x <- Just 3 Prelude| y <- Just "!" Prelude| Just (show x ++ y) Prelude| :} <interactive>:145:5: parse error on input `=' --------------- Any help would be appreciated. Thanks, Shishir

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Operator orders affect parse result
by m00nlight 13 Apr '15

13 Apr '15

Dear Haskellers, I have tried the "while language parser" tutorial on haskellwiki, but I encounter an problem I can not understand here, The following is the code: http://pastebin.com/saC9vwCn The problem is, with the code, the parse result of "fact := a / b * c" is *Main> parseString "fact := a / b * c" Assign "fact" (ABinary Multiply (ABinary Divide (Var "a") (Var "b")) (Var "c")) *Main> But if I swap the order of "/" and "*" operators in aOperators(make "*" appear before "/" operator), then the result is *Main> parseString "fact := a / b * c" Assign "fact" (ABinary Divide (Var "a") (ABinary Multiply (Var "b") (Var "c"))) *Main> So it seems that the order of the operator will affect the parse result. What's the problem with my code? How can I make my program to produce consistent parse result. Best Regards, --m00nlight

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Type declaration
by Christian Sperandio 13 Apr '15

13 Apr '15

Hi, I’m currently playing with the mutable array to implement an heap sort sort (from The Art of Computer Programming) and I’m puzzled with an error. When I code: toHeap :: Ord a => [a] -> IO [a] toHeap [] = return [] toHeap [x] = return [x] toHeap xs = do arr <- newListArray (1, length xs) xs :: IO (IOArray Int a) getElems arr I’ve got this error: Could not deduce (a ~ a1) from the context (Ord a) bound by the type signature for toHeap :: Ord a => [a] -> IO [a] at /var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:9:11-32 ‘a’ is a rigid type variable bound by the type signature for toHeap :: Ord a => [a] -> IO [a] at /var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:9:11 ‘a1’ is a rigid type variable bound by an expression type signature: IO (IOArray Int a1) at /var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:14:10 Expected type: [a1] Actual type: [a] Relevant bindings include xs :: [a] (bound at /var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:12:8) toHeap :: [a] -> IO [a] (bound at /var/folders/fk/339860r16j759wmsfmjkyr4h0000gn/T/flycheck797T5s/Heapsort.hs:10:1) In the second argument of ‘newListArray’, namely ‘xs’ In a stmt of a 'do' block: arr <- newListArray (1, length xs) xs :: IO (IOArray Int a) But with the code below, all work: toHeap :: Ord a => [a] -> IO [a] toHeap [] = return [] toHeap [x] = return [x] toHeap xs = do arr <- buildArray xs getElems arr buildArray :: [a] -> IO (IOArray Int a) buildArray xs = newListArray (1, length xs) xs What do I miss ? Thanks. Chris

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Linebreaks in lhs2tex eval
by martin 12 Apr '15

12 Apr '15

Hello all, it appears lhs2tex's \eval command does not show linebreaks. This is a pity, particularly when the output is formatted with a pretty printer. Is there a way to teach \eval to respect the linebreaks?

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Need help to write join on my Temporals
by martin 12 Apr '15

12 Apr '15

Hello all, I have these Types type Time = Integer data Change a = Chg { ct :: Time, -- "change time" cv :: a -- "change value" } deriving (Eq,Show) data Temporal a = Temporal { td :: a, -- "temporal default" tc :: [Change a] -- "temporal changes" } deriving (Eq, Show) And I am trying to make Temporal a Monad by implementing join. My first Attempt was awfully verbose. It was also faulty, which was revealed by this QuickQueck (I am happy with my implementation of <*>) prop_tJoin tpr1 tpr2 = let f = (*) y1 = f <$> tpr1 <*> tpr2 y2 = (f <$> tpr1) `ap` tpr2 in y1 == y2 While I understand why my first implementatin was faulty, I fail to come up with anything remotely readable. I don't understand why I am having such difficulties, let alone solve them. I feel I am approaching this from the wrong angle. Any advice would be very much appreciated.

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Debugging in Haskell
by Dimitri DeFigueiredo 12 Apr '15

12 Apr '15

I need to improve my Haskell debugging skills. I know of quickcheck, but that's for testing. It seems that: - Debug.Trace and - dynamic breakpoints in GHCi Are the two easy ways to check the state of your program at a specific point in execution. Is there another simple tool that I should know about? Any tips? Thank you, Dimitri

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Managing boilerplate code in QuickCheck properties...
by Stuart Hungerford 11 Apr '15

11 Apr '15

Hi, I'm using the wonderful QuickCheck and Tasty libraries to test a bunch of properties (actually axioms for typeclasses): positivity_suite :: TestTree positivity_suite = testGroup "positivity axiom" [ testProperty "Float" (axiom_positivity :: Float -> Bool), testProperty "(Float, Float)" (axiom_positivity :: (Float, Float) -> Bool), -- etc ] symmetry_suite :: TestTree symmetry_suite = testGroup "symmetry axiom" [ testProperty "Float" (axiom_symmetry :: Float -> Float -> Bool), testProperty "(Float, Float)" (axiom_symmetry :: (Float, Float) -> (Float, Float) -> Bool), -- etc ] Over multiple typeclasses, axioms and instance types this leads to a huge amount of repeated, boilerplate code. I asked an earlier question on the Haskell Cafe mailing list about testing polymorphic typeclasses in QuickCheck but I'm realizing the issue is going to crop up beyond testing typeclass axioms and perhaps beyond testing. I can see several workarounds for managing the boilerplate code: 1. Define a set of CPP macros for the property type signatures 2. Do the same sort of thing with Template Haskell but with mapping over a list of desired test types 3. Define a very tedious but succinct set of type aliases allowing most properties to become more cryptic one-liners type Pr_F_F = Float -> Float -> Bool type Pr_F2_F2 = (Float, Float) -> (Float, Float) -> Bool -- etc, then testProperty "(Float, Float)" (axiom_symmetry :: Pr_F2_F2) How do experienced Haskellers deal with this situation? I feel I'm missing something in how I've gone about this. Thanks, Stu

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Why does QuickCheck insist on this class constraint?
by martin 11 Apr '15

11 Apr '15

Hello all, Ghc complainend about Could not deduce (Arbitrary a1) arising from a use of ‘arbitrary’ Could not deduce (Eq a1) arising from a use of ‘arbitrary’ in the following code .. data CList a = CList [Change a] deriving Show instance (Arbitrary c, Eq c) => Arbitrary (CList c) where arbitrary = do ts <- orderedList :: Gen[Time] vs <- listOf arbitrary :: Arbitrary c => Gen [c] return $ CList $ zipWith (\t v -> Chg t v) (nub ts) vs instance (Arbitrary a, Eq a) => Arbitrary (Temporal a) where arbitrary = do d <- arbitrary (CList cs) <- arbitrary :: (Arbitrary a, Eq a) => Gen (CList a) -- <=== return (Temporal d cs) ..if I leave out the class constraints in the line maked with "<===". Why is that so, i.e. why isn't the constraint four lines up (in the instance declaration) sufficient?

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(.) vs ($)
by Vale Cofer-Shabica 06 Apr '15

06 Apr '15

Could someone please explain why the commented line fails spectacularly while the final line succeeds? >import System.IO (getContents) >import System.Environment (getArgs) >fileInput :: IO String >fileInput = getArgs>>=readFirst where > readFirst :: [FilePath] -> IO String > readFirst [] = System.IO.getContents >--readFirst xs = readFile.head xs > readFirst xs = readFile $ head xs I'm particularly confused given the following typings (from ghci): readFile.head :: [FilePath] -> IO String readFile.head [] :: a -> IO String And this is still stranger: :type readFile.head ["foo", "bar"] <interactive>:28:16: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "foo" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]' <interactive>:28:23: Couldn't match expected type `a0 -> FilePath' with actual type `[Char]' In the expression: "bar" In the first argument of `head', namely `["foo", "bar"]' In the second argument of `(.)', namely `head ["foo", "bar"]' Many thanks in advance, vale

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