- Beginners - Haskell.org

Resource-recommendation
by nikhil kaushik 14 Nov '17

14 Nov '17

Hello, I have just started learning Haskell and I am really enjoying it. As I do not have a good mathematical background and a little experience with programming in general, which book is recommended for me? I started learning haskell from LYAH website and I completed 6 chapters but there are no exercises in that book so I want a book which is beginner friendly and has exercises. Please recommended some books considering my situation. Thank you -- ~Nikhil Kaushik

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Type constructor
by mike h 08 Nov '17

08 Nov '17

Hi, I’m modelling fractions with a view to looking at continued fractions and I have this recursive structure. type Numerator = Integer data Fraction = Numbr Integer | F Numerator Fraction in ghci I do λ-> :t F 1 (Numbr 2) F 1 (Numbr 2) :: Fraction which is fine. But what surprised me is that it also works without using Numbr e.g. λ-> :t F 1 2 F 1 2 :: Fraction why is this? Thanks Mike

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Could not get parser ready
by Marcus Manning 06 Nov '17

06 Nov '17

Hello, I follow the instructions of script [1] in order to set up a parser functionality. But I' get into problems at page 202 with the code: p :: Parser (Char,Char) p = do x ← item item y ← item return (x,y) ghci and ghc throw errors: Prelude> let p:: Parser (Char,Char); p = do {x <- item; item; y <- item; return (x,y)} <interactive>:10:65: error: • Couldn't match type ‘[(Char, String)]’ with ‘Char’ Expected type: String -> [((Char, Char), String)] Actual type: Parser ([(Char, String)], [(Char, String)]) • In a stmt of a 'do' block: return (x, y) In the expression: do x <- item item y <- item return (x, y) In an equation for ‘p’: p = do x <- item item y <- item .... Did the semantics of do expr changed? [1] https://userpages.uni-koblenz.de/~laemmel/paradigms1011/resources/pdf/haske… Cheers, iconfly.

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Problem with Parser combinators
by Baa 03 Nov '17

03 Nov '17

Hello, all! I try this parser for `Read` instance: import qualified Text.ParserCombinators.ReadP as P import qualified Text.ParserCombinators.ReadPrec as RP import Data.Text as T instance Read RV where readPrec = parens $ do RP.lift P.skipSpaces s <- RP.look guard (not $ null s) (RP.+++) (RV1 <$> readPrec::RP.ReadPrec RV) (pure $ RV2 $ T.pack s) And see that `+++` (or <|>) works in strange way (I don't understand what happens). IMHO if first parser fails (i.e. `RV1 <$> readPrec::RP.ReadPrec RV`) then second should return `RV2 $ T.pack s`). But seems that if 1st fails then all fail too! I tried `+++`, `<++`, `<|>` - the same result. How is it possible?! I need to return something if parsing fails. Best is to return another parser but initial problem was that alternative parser must return the same type as first one. So, I decided to return "constant" - here it is `RV2 $ T.pack s`. Guarding from empty string is needed for this alternative value which I'll return when 1st parser fails. But... seems something is wrong. === Best regards, Paul

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How to write Read instance
by Baa 02 Nov '17

02 Nov '17

Hello all! I found some errors in the reading of previously shown big and complex record. Reason is: some of fields are reading wrongly. So, is there any useful documents how to write `Read` instance correctly (because I can't find good tutorial in the Web and often hit errors with `Read` instance)? May be tutorial/examples/any good info... === Best regards, Paul

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What's wrong with Haddock?!
by Baa 26 Oct '17

26 Oct '17

I often hit such errors: parse error on input ‘-- ^ my doc-string is here’ when generate Haddock documentation. This happens on line like: instance Something UTCTime where some a = blahBlah -- ^ my doc-string is here Sometimes such doc-strings are passing, sometimes - not. I can understand what is the reason. Is there some rule how to format such kind of doc-strings? === Cheers, Paul

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Re: [Haskell-beginners] Is it possible such Monad? (Baa)
by 鲍凯文 26 Oct '17

26 Oct '17

Hi, 1. I think your data type is isomorphic to data Allpass w m a = Pass Bool (m a) w 2. Maybe first try to write a Functor instance? Then you can find out what constraints are necessary for 'w' and 'm' to write it. 3. What does this data type intend to represent? Best, toz On Wed, Oct 25, 2017 at 5:32 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 it possible such Monad? (Baa) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 25 Oct 2017 13:56:31 +0300 > From: Baa <aquagnu(a)gmail.com> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: [Haskell-beginners] Is it possible such Monad? > Message-ID: <20171025135631.1f9913d2@Pavel> > Content-Type: text/plain; charset=US-ASCII > > Hello All! > > Is it possible to write Monad for such type: > > data Allpass w m a = Nopass (m a) w | Allpass (m a) w > > I can write (>>=), IMHO such type can not be Monad due to `w` is not > under `m` monad, right? > > > === > Best regards, Paul > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > > ------------------------------ > > End of Beginners Digest, Vol 112, Issue 22 > ****************************************** >

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Is it possible such Monad?
by Baa 25 Oct '17

25 Oct '17

Hello All! Is it possible to write Monad for such type: data Allpass w m a = Nopass (m a) w | Allpass (m a) w I can write (>>=), IMHO such type can not be Monad due to `w` is not under `m` monad, right? === Best regards, Paul

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Manipulate list, one element at a time
by martin 24 Oct '17

24 Oct '17

Hello all, How can I do something to each element of a list, but only modify one element at at time, so the result is a list of lists, where each sublist is the original list with one element altered. Something like type Each a = (a->a) -> [a] -> [[a]] I came up with: each :: Each a each f [] = [] each f (x:xs) = (f x : xs) : (map (x:) $ each f xs) λ> each (*10) [1..3] [[10,2,3],[1,20,3],[1,2,30]] but I wonder if there is a more standard way of doing this

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runState (liftM (+100) pop) [1, 2, 3, 4] (Example from LYH)
by Olumide 24 Oct '17

24 Oct '17

Dear List, Apologies for the awkward question title -- its the best I can do ATM. I'm still working my way through chapter 13 of LYH (title for a few monads more) and I came across the following two beasts ghci> runState (liftM (+100) pop) [1,2,3,4] (101,[2,3,4]) ghci> runState (fmap (+100) pop) [1,2,3,4] (101,[2,3,4]) See http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions Even though I'm still struggling to wrap my mind around monads, I sort of understand what's going on here. The problem is that I can't explain why the function (+100) is applied to _only_ the value 1 in (1,[2,3,4]). Regards, - Olumide

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