- Beginners - Haskell.org

how do typeclasses work again?
by Nicholls, Mark 12 Feb '17

12 Feb '17

Sorry..I do haskell about once every 6 months for 2 hours...and then get on with my life. I always forget some nuance of typeclasses. Consider some simple typeclass > class Is isx x where > apply :: (x -> y) -> isx -> y We can make any type a member of it...mapping to itself > instance Is x x where > apply f = f But we can also make a tuple a member of it...and pull the 1st member.. > instance Is (x,y) x where > apply f (x,y) = f x Weird and largey useless...but I'm playing. Then construct a function to operate on it > foo2 :: (Is isx Integer) => isx -> String > foo2 = apply (\i -> "") And... • Could not deduce (Is isx x0) arising from a use of ‘apply’ from the context: Is isx Integer bound by the type signature for: foo2 :: Is isx Integer => isx -> String at prop.lhs:51:3-43 The type variable ‘x0’ is ambiguous Relevant bindings include foo2 :: isx -> String (bound at prop.lhs:52:3) These potential instances exist: instance Is x x -- Defined at prop.lhs:41:12 instance Is (x, y) x -- Defined at prop.lhs:45:12 • In the expression: apply (\ i -> "") In an equation for ‘foo2’: foo2 = apply (\ i -> "") What's it going on about? (my brain is locked in F# OO type mode) I've told it to expect a function "Integer -> String"...surely? Whats the problem. CONFIDENTIALITY NOTICE This e-mail (and any attached files) is confidential and protected by copyright (and other intellectual property rights). If you are not the intended recipient please e-mail the sender and then delete the email and any attached files immediately. Any further use or dissemination is prohibited. While MTV Networks Europe has taken steps to ensure that this email and any attachments are virus free, it is your responsibility to ensure that this message and any attachments are virus free and do not affect your systems / data. Communicating by email is not 100% secure and carries risks such as delay, data corruption, non-delivery, wrongful interception and unauthorised amendment. If you communicate with us by e-mail, you acknowledge and assume these risks, and you agree to take appropriate measures to minimise these risks when e-mailing us. MTV Networks International, MTV Networks UK & Ireland, Greenhouse, Nickelodeon Viacom Consumer Products, VBSi, Viacom Brand Solutions International, Be Viacom, Viacom International Media Networks and VIMN and Comedy Central are all trading names of MTV Networks Europe. MTV Networks Europe is a partnership between MTV Networks Europe Inc. and Viacom Networks Europe Inc. Address for service in Great Britain is 17-29 Hawley Crescent, London, NW1 8TT.

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Re: [Haskell-beginners] how do typeclasses work again?
by Dimitri DeFigueiredo 11 Feb '17

11 Feb '17

I would like to make a general comment that is likely not useful to the specific question asked (which I think has already been nicely answered by others). But I would like to point out this video of an excellent lecture on typeclasses given by Simon Peyton-Jones a few years ago. https://channel9.msdn.com/posts/MDCC-TechTalk-Classes-Jim-but-not-as-we-kno… Cheers, Dimitri -- 2E45 D376 A744 C671 5100 A261 210B 8461 0FB0 CA1F

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Semigroup Instances
by Atrudyjane 07 Feb '17

07 Feb '17

I'm currently studying semigroups and trying to figure out how to determine which type variables need a semigroup instance. Here are a couple of examples from Evan Cameron's github (https://github.com/leshow/haskell-programming-book/blob/master/src/Ch15ex.hs) (1) data Validation a b = Failure a | Success b deriving (Eq, Show) instance Semigroup a => Semigroup (Validation a b) where Success a <> Success b = Success a Failure a <> Success b = Success b Success a <> Failure b = Success a Failure a <> Failure b = Failure (a <> b) * Why doesn't 'b' need an instance of semigroup? (2) newtype AccumulateRight a b = AccumulateRight (Validation a b) deriving (Eq, Show) instance Semigroup b => Semigroup (AccumulateRight a b) where AccumulateRight (Success a) <>AccumulateRight (Failure b) =AccumulateRight (Success a) AccumulateRight (Failure a) <>AccumulateRight (Success b) =AccumulateRight (Success b) AccumulateRight (Failure a) <>AccumulateRight (Failure b) =AccumulateRight (Failure a) AccumulateRight (Success a) <> AccumulateRight (Success b) = AccumulateRight (Success (a <> b)) * Why doesn't 'a' need an instance of semigroup? Thank you, Andrea Sent with [ProtonMail](https://protonmail.com) Secure Email.

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Applicative for State
by mike h 07 Feb '17

07 Feb '17

Hi, I have a State by another name, Stat, just to experiment and learn. newtype Stat s a = Stat { runStat :: s -> (a, s) } instance Functor (Stat s) where fmap f (Stat g) = Stat $ \s -> (f a, s) where (a, s) = g s instance Applicative (Stat s) where pure a = Stat $ \s -> (a, s) (Stat f) <*> (Stat g) = Stat $ \s -> (a, s) where (a, s) = undefined I really can’t get what the <*> in the Applicative should be! I just do see how I ‘get the f out of the Stat’ and then apply it. I’d be really grateful if someone would explain what it should be and the steps/reasoning needed to get there. Many thanks Mike

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Ambiguous type variable prevents the constraint `(Ord t0)' from being solved.
by Ivan Kush 06 Feb '17

06 Feb '17

I get this error (full message at the end of the mail). How could I correct my code? =================== Code: =================== module Intro where import Data.Bits -- for xor, .&. data Func a b = Empty | Leaf Int [(a, b)] | Branch Int Int (Func a b) (Func a b) applyd = let apply_listd l d x = case l of [] -> d x (a, b) : t -> let c = compare x a in if c == EQ then b else if c == GT then apply_listd t d x else d x in \f d x -> let k = 5 -- hash x - todo in let look t = case t of Leaf h l | h == k -> apply_listd l d x Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> -- (Branch p b l r) | ((k xor p) .&. (b - 1)) == 0 -> look (if k .&. b == 0 then l else r) _ -> d x in look f =================== Error: =================== Intro.hs:37:25: error: * Ambiguous type variable `t0' arising from a use of `apply_listd' prevents the constraint `(Ord t0)' from being solved. Relevant bindings include l :: [(t0, t)] (bound at Intro.hs:36:28) t :: Func t0 t (bound at Intro.hs:34:21) look :: Func t0 t -> t (bound at Intro.hs:34:16) x :: t0 (bound at Intro.hs:32:15) d :: t0 -> t (bound at Intro.hs:32:13) f :: Func t0 t (bound at Intro.hs:32:11) (Some bindings suppressed; use -fmax-relevant-binds=N or -fno-max-relevant-binds) Probable fix: use a type annotation to specify what `t0' should be. These potential instances exist: instance Ord Ordering -- Defined in `GHC.Classes' instance Ord Integer -- Defined in `integer-gmp-1.0.0.1:GHC.Integer.Type' instance Ord a => Ord (Maybe a) -- Defined in `GHC.Base' ...plus 22 others ...plus five instances involving out-of-scope types (use -fprint-potential-instances to see them all) * In the expression: apply_listd l d x In a case alternative: Leaf h l | h == k -> apply_listd l d x In the expression: case t of { Leaf h l | h == k -> apply_listd l d x Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> look (if k .&. b == 0 then l else r) _ -> d x } -- Best wishes, Ivan Kush

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Need help understanding the tell function in the Monad Writer example in LYAH
by Olumide 06 Feb '17

06 Feb '17

Hello List, Would someone kindly explain how the tell function interacts with the Writer Monad as show below: (Example taken from chapter 14 of LYAH http://learnyouahaskell.com/for-a-few-monads-more#reader ) multWithLog :: Writer [String] Int multWithLog = do a <- logNumber 3 b <- logNumber 5 tell ["Gonna multiply these two"] return (a*b) Result: ghci> runWriter multWithLog (15,["Got number: 3","Got number: 5","Gonna multiply these two"]) I know that tell function binds to an argument that is discarded but I don't know how its argument "Gonna multiply these two" is concatenated with the other the Writer created by logNumber 3 and logNumber 5. Also, I don't understand the paragraph following the example: "It's important that return (a*b) is the last line, because the result of the last line in a do expression is the result of the whole do expression. Had we put tell as the last line, () would have been the result of this do expression. We'd lose the result of the multiplication. However, the log would be the same." Regards, - Olumide

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Bit confused...
by mike h 04 Feb '17

04 Feb '17

I have ---------- import qualified Data.Map as M type Link a = (a, Int) data MChain a = Map a [Link a] deriving (Show) ------------------- and want to make a Monoid of MChain. So I have ------------------- instance Monoid (MChain a) where mempty = M.empty mappend = undefined ------------------- this won’t compile and I need M.empty to be Map a [Link a] The error is Couldn't match expected type ‘MChain a1’ with actual type ‘M.Map k0 a0’ • In the expression: M.empty :: MChain a In an equation for ‘mempty’: mempty = M.empty :: MChain a In the instance declaration for ‘Monoid (MChain a)’ This part confuses me "Couldn't match expected type ‘MChain a1 with actual type ‘M.Map k0 a0’ " How should I proceed. Many thanks Mike

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Question about example in 'using do notation with Writer' section of LYH
by Olumide 03 Feb '17

03 Feb '17

Hello List, Chapter 14 of LYH ( http://learnyouahaskell.com/for-a-few-monads-more#reader ) has the following bit of code: import Control.Monad.Writer logNumber :: Int -> Writer [String] Int logNumber x = Writer (x, ["Got number: " ++ show x]) multWithLog :: Writer [String] Int multWithLog = do a <- logNumber 3 b <- logNumber 5 return (a*b) I have a fair grasp of what's going bu the last line return(a*b) eludes me. Specifically its the a*b part that baffles me. I think 3 is bound to 'a' and 5 to 'b', then return(a*b) would put the value 15 in a context ... So what becomes of the string parts of the Writer monads created by logNumber 3 and logNumber 5 respectively. Regards, - Olumide

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applicative instance
by sasa bogicevic 30 Jan '17

30 Jan '17

What is wrong with my applicative instance for custom List type ? http://lpaste.net/351723 data List a = Nil | Cons a (List a) deriving (Eq, Show) instance Applicative List where pure x = Cons x Nil Nil <*> _ = Nil _ <*> Nil = Nil (Cons x xy) <*> (Cons z dy) = Cons (x z) (xy <*> dy) Prelude> let functions = Cons (+1) (Cons (*2) Nil) Prelude> let values = Cons 1 (Cons 2 Nil) Prelude> functions <*> values Cons 2 (Cons 3 (Cons 2 (Cons 4 Nil))) -- I get Cons 2 (Cons 4 Nil) what is wrong with my Applicative instance ? { name: Bogicevic Sasa phone: +381606006200 }

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Re: [Haskell-beginners] Ambiguous type variable prevents the constraint `(Ord t0)' from being solved.
by 鲍凯文 30 Jan '17

30 Jan '17

Hi, I refactored it as such: apply_listd [] d x = d x apply_listd ((a,b):t) d x = case compare x a of EQ -> a GT -> b _ -> d x applyd f d x = look f where k = 5 look (Leaf h l) | h == k = apply_listd l d x look (Branch p b l r) | (k `xor` p) .&. (b - 1) == 0 = look (if k .&. b == 0 then l else r) look _ = d x ...and it compiled so i'm not sure what happened there. Simpler though: adding the type signature applyd :: Ord a => Func a b -> (a -> b) -> a -> b also made it compile. I'm not sure why ghc thinks there's multiple instances to choose from (somebody please explain), but in any case adding type signatures will give you better error messages/help ghc. Hope that helps a bit. On Sun, Jan 29, 2017 at 4:00 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. Ambiguous type variable prevents the constraint `(Ord t0)' > from being solved. (Ivan Kush) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Sat, 28 Jan 2017 23:09:09 +0300 > From: Ivan Kush <ivan.v.kush(a)yandex.ru> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Ambiguous type variable prevents the > constraint `(Ord t0)' from being solved. > Message-ID: <5152611485634149(a)web34j.yandex.ru> > Content-Type: text/plain; charset=utf-8 > > I get this error (full message at the end of the mail). How could I > correct my code? > > > =================== > Code: > =================== > > module Intro where > > import Data.Bits -- for xor, .&. > > data Func a b > = Empty > | Leaf Int [(a, b)] > | Branch Int Int (Func a b) (Func a b) > > applyd = > let apply_listd l d x = > case l of > [] -> d x > (a, b) : t -> > let c = compare x a > in if c == EQ then b > else if c == GT then apply_listd t d x > else d x > > in \f d x -> > let k = 5 -- hash x - todo > in let look t = > case t of > Leaf h l | h == k -> > apply_listd l d x > Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> -- > (Branch p b l r) | ((k xor p) .&. (b - 1)) == 0 -> > look (if k .&. b == 0 then l else r) > _ -> d x > in look f > > > > =================== > Error: > =================== > > Intro.hs:37:25: error: > * Ambiguous type variable `t0' arising from a use of `apply_listd' > prevents the constraint `(Ord t0)' from being solved. > Relevant bindings include > l :: [(t0, t)] (bound at Intro.hs:36:28) > t :: Func t0 t (bound at Intro.hs:34:21) > look :: Func t0 t -> t (bound at Intro.hs:34:16) > x :: t0 (bound at Intro.hs:32:15) > d :: t0 -> t (bound at Intro.hs:32:13) > f :: Func t0 t (bound at Intro.hs:32:11) > (Some bindings suppressed; use -fmax-relevant-binds=N or > -fno-max-relevant-binds) > Probable fix: use a type annotation to specify what `t0' should be. > These potential instances exist: > instance Ord Ordering -- Defined in `GHC.Classes' > instance Ord Integer > -- Defined in `integer-gmp-1.0.0.1:GHC.Integer.Type' > instance Ord a => Ord (Maybe a) -- Defined in `GHC.Base' > ...plus 22 others > ...plus five instances involving out-of-scope types > (use -fprint-potential-instances to see them all) > * In the expression: apply_listd l d x > In a case alternative: Leaf h l | h == k -> apply_listd l d x > In the expression: > case t of { > Leaf h l | h == k -> apply_listd l d x > Branch p b l r > | (k `xor` p) .&. (b - 1) == 0 > -> look (if k .&. b == 0 then l else r) > _ -> d x } > > > -- > Best wishes, > Ivan Kush > > > ------------------------------ > > Subject: Digest Footer > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > > ------------------------------ > > End of Beginners Digest, Vol 103, Issue 25 > ****************************************** >

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