November 2015 - Beginners

Learning Haskell: Higher-order Functions
by Manuel M T Chakravarty 18 Nov '15

18 Nov '15

We have got a new tutorial chapter on ”Higher-order Functions & Combinators”: http://blog.haskellformac.com/blog/higher-order-functions-combinators Hope you enjoy it! Manuel

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Applicative Laws for the ((->) r) type
by Umair Saeed 18 Nov '15

18 Nov '15

Hi all, I'm trying to check that the Applicative laws hold for the function type ((->) r), and here's what I have so far: -- Identiy pure (id) <*> v = v pure (id) <*> v const id <*> v (\x -> const id x (g x)) (\x -> id (g x)) (\x -> g x) g x v -- Homomorphism pure f <*> pure x = pure (f x) pure f <*> pure x const f <*> const x (\y -> const f y (const x y)) (\y -> f (x)) (\_ -> f x) pure (f x) Did I perform the steps for the first two laws correctly? I'm struggling with the interchange & composition laws. For interchange, so far I have the following: -- Interchange u <*> pure y = pure ($y) <*> u u <*> pure y u <*> const y (\x -> g x (const y x)) (\x -> g x y) -- I'm not sure how to proceed beyond this point. I would appreciate any help for the steps to verify the Interchange & Composition applicative laws for the ((->) r) type. Thank you, ~Umair

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Re: [Haskell-beginners] Is there an idiom for this?
by 鲍凯文 16 Nov '15

16 Nov '15

I think i saw a relevant stackoverflow question before, but I can't find it. Anyways, the answer said something like: liftA2 (&&) pred1 pred2 which has the type a -> Bool, as desired. The reader applicative gives your input to both predicates and (&&) is applied to their results. Hope I remembered this correctly. On Nov 16, 2015 4:01 AM, <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Is there an idiom for this? (Mark Carter) > 2. Re: Is there an idiom for this? (emacstheviking) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 16 Nov 2015 11:44:35 +0000 > From: Mark Carter <alt.mcarter(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Is there an idiom for this? > Message-ID: <5649C1A3.1010901(a)gmail.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Suppose I want to use an argument twice, as for example in the expression: > (\x -> (pred1 x) and (pred2 x)) > > Is there a shorter way of doing this? > > > ------------------------------ > > Message: 2 > Date: Mon, 16 Nov 2015 11:53:28 +0000 > From: emacstheviking <objitsu(a)gmail.com> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: Re: [Haskell-beginners] Is there an idiom for this? > Message-ID: > <CAEiEuULHWhechDwUF417Rnr55RDrEVUFVOgiQ6= > 5-8jux3pXyg(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > I guess it depends on the final use cases... you could use currying to > partially evaluate some stuff ready, locked and loaded as it were but the > example you have given shows to distinct functions pres1 and pred2. > > I guess the short answer is "yes" but it depends on how you do it! > > :) > Sean > > > On 16 November 2015 at 11:44, Mark Carter <alt.mcarter(a)gmail.com> wrote: > > > Suppose I want to use an argument twice, as for example in the > expression: > > (\x -> (pred1 x) and (pred2 x)) > > > > Is there a shorter way of doing this? > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > >

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Why this order of parameters
by martin 16 Nov '15

16 Nov '15

Hello all, there are two functions, where the order of parameters confuses me. Ons is runState :: State s a -> s -> (a, s) I understand that in the constructor s has to be first, so we can turn (State s) into a monad. But why doesn't s come first in the result too, as in runState :: State s a -> s -> (s, a) The other example is foldl, foldr foldl :: (a -> b -> a) -> a -> [b] -> a foldr :: (a -> b -> b) -> b -> [a] -> b For once the list is a list of as in foldl, but a list of as in foldr. Now that can be fixed with renaming the type parameters foldr :: (b -> a -> a) -> a -> [b] -> a this is the exact same thing and resembles more the type of foldl. But still the function argument has its parameters flipped. foldl's function takes the list element as second parameter and foldr's takes it as first parameter. Why is that so?

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Is there an idiom for this?
by Mark Carter 16 Nov '15

16 Nov '15

Suppose I want to use an argument twice, as for example in the expression: (\x -> (pred1 x) and (pred2 x)) Is there a shorter way of doing this?

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which Applicative instance?
by Graham Gill 15 Nov '15

15 Nov '15

I'm following the CIS194 lectures and exercises. In particular, in one of the Applicative lectures [0] we're asked to implement mapA and sequenceA, from the name and given type signature alone. I renamed them mapA' and sequenceA' to avoid name clashes, and defined: import Control.Applicative mapA' :: Applicative f => (a -> f b) -> ([a] -> f [b]) mapA' f = sequenceA' . map f sequenceA' :: Applicative f => [f a] -> f [a] sequenceA' = foldr (\fa fla -> (:) <$> fa <*> fla) (pure []) These implementations seem to be correct, though undoubtedly not the nicest. Now, I ask for a type in ghci: > :t mapA' pure [1..5] mapA' pure [1..5] :: (Enum b, Num b, Applicative f) => f [b] And so I try the following: > mapA' pure [1..5] :: [[Int]] [[1,2,3,4,5]] > take 5 . getZipList $ (mapA' pure [1..5] :: ZipList [Int]) [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]] > mapA' pure [1..5] :: Maybe [Int] Just [1,2,3,4,5] That makes sense. But I don't understand the following: > mapA' pure [1..5] [1,2,3,4,5] Which Applicative instance is being used here? The type test above says my expression should have type f [b], where f is an Applicative and b is a Num, but all I see is a result of type [b] when I compute mapA' pure [1..5]. What's going on? More simply > :t pure pure :: Applicative f => a -> f a > pure [1..5] [1,2,3,4,5] a must match a list of Nums, but then where did f go? Is ghci using some default instance? Actually, after I add the following lines to my source file: ggg = pure [1..5] hhh = mapA' pure [1..5] I get the errors: No instance for (Applicative f0) arising from a use of ‘pure’ No instance for (Applicative f1) arising from a use of ‘mapA'’ respectively. That makes sense to me. Graham [0] http://www.seas.upenn.edu/~cis194/fall14/spring13/lectures/11-applicative2.…

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Query regarding an unusually behaving code
by Abhishek Kumar 14 Nov '15

14 Nov '15

Bellow is my code for returning a list of digits given an integer.I'm unable to find any bug in it but somehow it crashes any computer I run it on when I try to calculate digs 100.Please tell me bug in my code. let digs 0 =[0] let digs x = (digs (x `div` 10)) ++ [(x `rem` 10)] digs 100 Thanks Abhishek

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Generate a list of all constructors
by Kees Bleijenberg 14 Nov '15

14 Nov '15

In this piece of code data Color = RED | BLUE | YELLOW allColors :: [Color] allColors = [RED, BLUE, YELLOW] I have to remeber that if I add a Color, I have to add the color in allColors too. Is there no better way to add all possible colors in allColors? Kees

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Opaque (to me) compilation error
by Dan Stromberg 14 Nov '15

14 Nov '15

In the following code: prefix_md5 :: String -> Data.ByteString.ByteString prefix_md5 filename = do let prefix_length = 1024 file <- System.IO.openBinaryFile filename System.IO.ReadMode :: (IO System.IO.Handle) data_read <- Data.ByteString.hGet file prefix_length :: (IO Data.ByteString.ByteString) _ <- System.IO.hClose file let hasher = Crypto.Hash.MD5.init :: Crypto.Hash.MD5.Ctx let hasher2 = Crypto.Hash.MD5.update hasher data_read :: Crypto.Hash.MD5.Ctx let digest = Crypto.Hash.MD5.finalize hasher2 :: Data.ByteString.ByteString return digest :: (IO Data.ByteString.ByteString) I get the error: Md5s.hs:13:5: Couldn't match type `IO Data.ByteString.ByteString' with `Data.ByteString.ByteString' Expected type: IO System.IO.Handle -> (System.IO.Handle -> IO Data.ByteString.ByteString) -> Data.ByteString.ByteString Actual type: IO System.IO.Handle -> (System.IO.Handle -> IO Data.ByteString.ByteString) -> IO Data.ByteString.ByteString In a stmt of a 'do' block: file <- System.IO.openBinaryFile filename System.IO.ReadMode :: IO System.IO.Handle How should I interpret that error to solve this kind of problem on my own in the future? I don't see where the line in question does anything with ByteString's! How might I correct this function to eliminate the error? Thanks! -- Dan Stromberg

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is this not a better way ? data structure
by Roelof Wobben 13 Nov '15

13 Nov '15

Hello, In the book there is a example of this data-structure. data Shape = Circle float float | Rectangle float float Now I have to change it to add a center point. But I wonder if this is not a better way to describe it Data Schape = Circle Radius | Rectangle Width Height Type Radius = Float type Width = Float type Height = Float Roelof

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