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Re: [Haskell-beginners] Fwd: Re: Multiple letters between -> ->
by Francesco Ariis 30 Nov '17

30 Nov '17

On Thu, Nov 30, 2017 at 12:55:07PM +0100, Marcus Manning wrote: > But Either is not (*->*)->*, instead it is *->(*->*). Is (*->*)->* > expressable in Haskell functions? Yes, with an appropriate data declaration: λ> data FooType a = FooCons (a Int) λ> :k FooType FooType :: (* -> *) -> *

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Fwd: Re: Multiple letters between -> ->
by Marcus Manning 30 Nov '17

30 Nov '17

-------- Forwarded Message -------- Subject: Re: [Haskell-beginners] Multiple letters between -> -> Date: Sat, 25 Nov 2017 13:05:25 +0100 From: Marcus Manning <iconsize(a)gmail.com> To: Francesco Ariis <fa-ml(a)ariis.it> Sorry, for the long break. Thanks for replying. I do not believe that h is a higher kinded type. What I want to express is that a function f could take a type constructor as argument and simply returns it, but f Maybe throws an Error <interactive>:13:3: error: • Data constructor not in scope: Maybe :: h a • Perhaps you meant variable ‘maybe’ (imported from Prelude) So what instead does h a mean in a function declaration? Cheers, Marcus. On 11/23/2017 06:27 PM, Francesco Ariis wrote: > On Thu, Nov 23, 2017 at 06:19:51PM +0100, Marcus Manning wrote: >> Hi, >> >> Original I thought a Signature like: >> >> f :: h a -> h a >> >> means that h is a higher kinded type just like in Type Classes ( for >> instance f in Functor f). >> >> But I heard such a meaning is not allowed in normal Haskell functions. What >> instead is the meaning of h a? > Hello Marcus, > you can write that but, since we know nothing about `h` and `a`, > the only possible (non-undefined) function to implement that > signature is: > > f :: h a -> h a > f = id > > Any other implementation would require us to know something about h, > hence a typeclass-constraint (e.g. Functor h =>) on h. > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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Type declarations
by Patrik Iselind 27 Nov '17

27 Nov '17

Hi, What's the difference between `delta :: (Point t) => t -> t -> Double` and `delta :: Point p -> Point q -> Double`. The later one is accepted by GHCI when i do :load. As i see it the first one would make better sense to me. I want two Point as in-parameters and delta will produce a Double, but GHCI refuse this saying ``` nine.hs:10:11: Expected a constraint, but ‘Point t’ has kind ‘*’ In the type signature for ‘delta’: delta :: (Point t) => t -> t -> Double Failed, modules loaded: none. ``` Is this saying that Point t match 'everything' (*)? In the second version, which is accepted by GHCI, i don't see the point of p and q. Can i use these somehow? All of delta using the accepted type declaration looks like this for reference: ``` data Direction d = LEFT | RIGHT | STRAIGHT deriving (Show) data Point a = Coordinate Double Double deriving (Show) -- Calculate the slope between two points (dy/dx) delta :: Point p -> Point q -> Double delta (Coordinate a b) (Coordinate c d) | (a == c) = 0 | otherwise = (d-b)/(c-a) angle (Coordinate g h) (Coordinate i d) (Coordinate e f) | (delta a b) > (delta b c) = RIGHT | (delta a b) < (delta b c) = LEFT | otherwise = STRAIGHT where a = Coordinate g h b = Coordinate i d c = Coordinate e f ``` I'm also wondering if there is a simpler way than recreating the Coordinate as a, b, and c in angle. It seems to work ok to me, i just feel that a, b, and c in angle should be possible to express in a better way. -- Patrik Iselind

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Why do i need to specify the class of a here at all?
by Patrik Iselind 26 Nov '17

26 Nov '17

Hi, I'm following the Real World Haskell book and in chapter three came across the exercise to sort a list of lists. I attach what i have so far. My question is the following. Why do i need to add "Eq a =>" to the type definition of sortListOfLists? The type of a should be irrelevant as i see it, my function couldn't care less what the type of the contents of the lists in the list are, or? All my function cares about is the length of those lists in the list. The designator a in the type declaration for sortListOfLists denote the type of the contents in the lists of lists as i understand it. So a list of lists containing object i can test for equality. Another thought i have is if i really have to specify that i expect list of lists. Couldn't that be list of something else with a length, a string perhaps, just as well? In other words a type like sortListOfLists :: a -> a. As i see it that should be just as valid as the type of what i return is the same as that which i get as input regardless of it being a list or not. -- Patrik Iselind

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How to use Text.Printf w/ custom types
by Baa 24 Nov '17

24 Nov '17

Hello All! As I understand to use Text.Printf.printf with custom types or to return result as Text instead of String I must implement some class instances: PrintfType, PrintfArg. But how to do this if they are exported from module as type-names only - without its methods (i.e. not "Printf (..)", but "Printf") ? I tried to import Text.Printf.Internal but seems that no such submodule... Any ideas? === Best regards, Paul

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Multiple letters between -> ->
by Marcus Manning 23 Nov '17

23 Nov '17

Hi, Original I thought a Signature like: f :: h a -> h a means that h is a higher kinded type just like in Type Classes ( for instance f in Functor f). But I heard such a meaning is not allowed in normal Haskell functions. What instead is the meaning of h a? Cheers, Marcus.

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mail.haskell.org Archive search
by trent shipley 21 Nov '17

21 Nov '17

Are the Haskell mail archives searchable? If so how? Trent

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Programming with Singletons
by Quentin Liu 18 Nov '17

18 Nov '17

Hi all, I was doing the “Singletons” problem at codewars[1]. The basic idea is to use dependent types to encode the length of the vector in types. It uses data Nat = Zero | Succ Nat data SNat a where SZero :: SNat Zero SSucc :: SNat a -> SNat (Succ a) to do the encoding. The vector is defined based on the natural number encoding: data Vec a n where VNil :: Vec a Zero VCons :: a -> Vec a n -> Vec a (Succ n) There are some type families declared for manipulating the natural numbers, and one of them that is relevant to the question is type family (Add (a :: Nat) (b :: Nat)) :: Nat type instance Add Zero b = b type instance Add a Zero = a type instance Add (Succ a) (Succ b) = Succ (Succ (Add a b)) where the `Add` function adds natural numbers. The problem I am stuck with is the concatenation of two vectors: (++) :: Vec v m -> Vec v n -> Vec v (Add m n) VNil ++ b = b (VCons x xs) ++ b = VCons x $ xs ++ b The program would not compile because the compiler found that `VCons x $ xs ++ b`gives type `Vec v (Succ (Add n1 n))`, which does not follow the declared type `Vec v (Add m n)`. Is it because ghc does not expand `Add m n’ that the type does not match? I read Brent Yorgey’s blog on type-level programming[2] and he mentioned that would not automatically expand types. But the posted time of the blog is 2010 and I am wondering if there is any improvement to the situation since then? Besides, what would be the solution to this problem Warm Regards, Qingbo Liu [1] https://www.codewars.com/kata/singletons/train/haskell [2] https://byorgey.wordpress.com/2010/07/06/typed-type-level-programming-in-ha…

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Non-exhaustive patterns
by Jim 18 Nov '17

18 Nov '17

Hey guys, Thiis is my function below: notThe :: String -> Maybe String notThe word | word /= "the" = Just word | word == [] = Just [] | otherwise = Nothing replaceThe :: String -> String replaceThe word = go (words word) where go (x:xs) | notThe x == Just [] = [] | notThe x == Just word = word ++ go xs | notThe word == Nothing = " a " ++ go xs > replaceThe "what" "what*** Exception: chap12.hs:(13,13)-(16,55): Non-exhaustive patterns in function go I thought I covered all the potential patterns in my replaceThe function. I'm not sure what pattern I've missed, any thoughts? Best regards, Jim

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argument counts...
by Gregory Guthrie 17 Nov '17

17 Nov '17

I have a simple function: revFn (x:xs) = (revFn xs) . (x:) Of course GHCi correctly infers the type as: revFn :: [a] -> [a] -> c Adding the base case: revFn [] xs = xs Now gives an error; "Equations for 'revF' have different numbers of arguments" Of course this can be "fixed" by either adding the cancelled argument to the first clause, or converting the base case to only have one explicit argument, and a RHS of a lambda or identity function. But since the interpreter already correctly inferred that the first clause has two arguments (with only one explicit), why does it then ignore this and give an error when the second clause shows two explicit arguments? The types are all correct in either case - why require explicit arguments? Or, perhaps I am missing something simple?

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