August 2017 - Beginners

Fractional of DiffTime
by Baa 18 Aug '17

18 Aug '17

Hello, List! DiffTime type has "closed" on itself operation "/". So, (/) :: DiffTime -> DiffTime -> DiffTime But from a physics point of view, it doesn't make sense. When you divide hours by hours you get "dimensionless" value, not units (hours) but simple number (I'm not sure how it's named in English). And it's true for any physical values: [m/s] : [m/s] = [Int] (not [m/s] sure) Is it right to define CLOSED "/" for such set like DiffTime? === Best regards, Paul

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Applicative: how <*> really works
by Yassine 17 Aug '17

17 Aug '17

Hi, I have a question about functor applicate. I know that: pure (+1) <*> Just 2 produce: Just 3 because pure (+1) produce Just (+1) and then Just (+1) <*> Just 2 produce Just (2+1) but in more complex case like: newtype Parser a = P (String -> [(a,String)]) parse :: Parser a -> String -> [(a,String)] parse (P p) inp = p inp item :: Parser Char item = P (\inp -> case inp of [] -> [] (x:xs) -> [(x,xs)]) instance Functor Parser where fmap g p = P (\inp -> case p inp of [] -> [] [(v, out)] -> [(g v, out)]) instance Applicative Parser where pure v = P (\inp -> [(v, inp)]) pg <*> px = P (\inp -> case parse pg inp of [] -> [] [(g, out)] -> parse (fmap g px) out) When I do: parse (pure (\x y -> (x,y)) <*> item <*> item) "abc" The answer is: [(('a','b'),"c")] But I don't understand what exactly happens. First: pure (\x y -> (x,y)) => P (\inp -> [(\x y -> (x,y), inp)]) Then: P (\inp -> [(\x y -> (x,y), inp)]) <*> item => ??? Can someone explain what's happens step by step please. Thank you.

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Re: [Haskell-beginners] Ambiguous type variable
by Jonathon Delgado 17 Aug '17

17 Aug '17

Thank you very much, that does clarify things! Francesco Ariis wrote: That's correct! Indeed ghc is not complaining about a lack of instances, as it would with, say λ> putStrLn 5 -- • No instance for (Num String) arising from etc etc. but about the *ambiguity* of type variable `e`. How does `catch` know _which_ exception to deal with if you don't specify the concrete type? Consider: prova = catch (print $ div (error "hey bby") 0) (\e -> print "ouch" >> print (e :: ErrorCall)) -- I want to deal with arithmetic errors here and not in `prova` palla = catch prova (\e -> print "baaah" >> print (e :: ArithException)) If I switch ArithException and ErrorCall the behaviour of invoking `palla` changes. Having a catch-all `catch` is possible by using (e :: SomeException); if you don't care about `e` and just want to do an action regardless, you are probably better off with `onException`. Ask more if needed!

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Re: [Haskell-beginners] Ambiguous type variable
by Jonathon Delgado 17 Aug '17

17 Aug '17

I'm sure it makes sense! I'm not really following though. I understood typeclasses to be analogous to OO interfaces. So if a variable implements the Exception interface, and Exception implements the Show interface, then it should automatically support show. I take it this was wrong? How does the compiler use typeclasses if they're not interfaces? Francesco Ariis wrote: > I'm trying to use > catch (...) (\e -> putStrLn $ show e) > However, I get an error > Ambiguous type variable ‘a0’ arising from a use of ‘show’ prevents the constraint ‘(Show a0)’ from being solved. > This goes away if I change the code to > catch (...) (\e -> putStrLn $ show (e::IOException)) > > A couple of things I don't understand here: > - The signature for catch begins "Exception e", and exception it "class (Typeable e, Show e) => Exception e". So why isn't show automatically available? > - Why does the new code work at all? e is Exception, not IOException. What would happen if it caught a different Exception? IOException is a concrete type while Exception is a typeclass. In the end, the compiler needs the former, the latter not being enough. The code works as any other class-based function would someFunction :: Monoid a -> [a] -> a -- ^-- in the end `Monoid a` will become something concrete, like -- a String, a Sum, etc. Does that make sense?

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Ambiguous type variable
by Jonathon Delgado 17 Aug '17

17 Aug '17

I'm trying to use catch (...) (\e -> putStrLn $ show e) However, I get an error Ambiguous type variable ‘a0’ arising from a use of ‘show’ prevents the constraint ‘(Show a0)’ from being solved. This goes away if I change the code to catch (...) (\e -> putStrLn $ show (e::IOException)) A couple of things I don't understand here: - The signature for catch begins "Exception e", and exception it "class (Typeable e, Show e) => Exception e". So why isn't show automatically available? - Why does the new code work at all? e is Exception, not IOException. What would happen if it caught a different Exception?

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How to "cabal install" on current Arch/Manjaro ?
by lists0＠freea2a.de 13 Aug '17

13 Aug '17

Helle dear readers, at the moment I'm not able to compile the most things with cabal/ghc. For example: cabal install cpphs gives Linking dist/build/cpphs/cpphs ... /usr/bin/ld: cannot find -lHStext-1.2.2.2-3ENqlljngKa6xj1Go2fVWq I have read that Arch linux has changed haskell to use dynamic linking, and I have installed the new packet "ghc-static" with static libraries. How can I switch between static and dynamic linking in cabal and what are the preconditions? Where can I give the directories for searching for haskell-libraries? Thank you in advance for help. Jürgen

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How to make instance of MonoFunctor
by Baa 10 Aug '17

10 Aug '17

Hello, Dear List! I want to make functor of my type like newtype UserName = UserName { unName :: Text } sure, it's impossible, so I will make MonoFunctor instead of (from library mono-traversable). So, I try: instance MonoFunctor UserName where omap fn (UserName n) = UserName $ fn $ n but I get error • Couldn't match expected type ‘Element UserName’ with actual type ‘Text’ • In the second argument of ‘($)’, namely ‘unName an’ In the second argument of ‘($)’, namely ‘fn $ unName an’ In the expression: UserName $ fn $ unName an interesting is that Text "type Element Text" of "type family Element mono" (instance?). So, how to make mono-functor for such `UserName` structure? -- Best regards, Paul

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Grappling with State Monad get
by Olumide 08 Aug '17

08 Aug '17

Ahoy Haskellers, In the section "Getting and Setting State" (http://learnyouahaskell.com/for-a-few-monads-more#state) in LYH get is defined as get = state $ \s -> (s, s) How does does get determine the type s, is considering that it has no argument as per the definition given above? Or is the definition written in some sort of point-free notation where an argument has been dropped? I find the line stackNow <- get in the the function(?) stackyStack confusing for the same reason. I guess my difficulty is that state $ \s ->(s , s) has a generic type (s) whereas the stackyStack has a concrete type. Is the type of s determined from the type of the stateful computation/do notation? Regards, - Olumide

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Password Program Questions
by Casey Newsome 03 Aug '17

03 Aug '17

I am very new to Haskell and I am attempting to make a simple password program as my first program. I have ran into a problem and I am not very sure on how to fix it. I do not know how to make a second section for the passwords. My guessing attempts have failed so I am asking for help. Here is my code main = do putStrLn "Hello, Who are you?" name <- getLine --User Input putStrLn ("Hey " ++ name ++ ", What's the password?") pass <- getLine if pass == "12345" then putStrLn ("Welcome") pw else do putStrLn "That is wrong!" mainpw = do putStrLn "What Password do you need?" Thanks in advance, Casey :D

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Libraries for Playing Audio
by Atrudyjane 02 Aug '17

02 Aug '17

Hello Cafe, Is there a recommended Haskell library for playing/manipulating audio files? There's a long list of sound libraries on Hackage, and was thinking looking into either conduit-audio or SDL-mixer. Thank You, Andrea Sent with [ProtonMail](https://protonmail.com) Secure Email.

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