November 2017 - Beginners

how to undertand the function join (*)
by Ray 30 Nov '17

30 Nov '17

Hello Marcus, what does join (*) do? (*) :: Num a => a -> a -> a join (*) :: Num a => a -> a when we feed a number to join (*),for instance; λ> : join (*) 3 9 it seems thata join (*) become a square function. what does join do to (*) to make that happen?

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How to parse hetero-list
by Baa 30 Nov '17

30 Nov '17

Hello, All! I have types A, B, C... They have instances of class Read, Show, SomethingElse, ... I want to keep them as one collection. For example, I can use such datatype for it: data Tags = forall a. Read a => Tags [a] Now I want to parse "some-string" into A or B or C or ... but without to create type's sum of A|B|C|...! All those types have `Read` instance but Haskell will not find what instance to use, right? How to parse string with alternatives where each alternative returns `ReadP A` or `ReadP B`, etc? My intuition is that I should make `x <|> y <|> z <|> ...` where x,y,z - parses string into A, B, C, ... so they are expressions like `readPrec :: ReadPrec A`, etc. But how to pass types into such expression? I must to make expression folding (with `<|>`) `readPrec::someType`, where `someType` is item of types list `{A, B, C, ...}` May be it's possible to be done with HList-like types, like: HSet, HVect or similar, but I don't know: - how to iterate/fold over types - I'm not sure that HSet/HVect keeps types (like `A ::: B ::: C ::: ...`) My idea is to have some function which can parse string into one of A|B|C|... with call like: myParse "some-string" :: (Something A ::: B ::: C ::: D) (suppose `:::` is type-level "cons"). So, `myParse` is polymorphic and it can fold/iterate over type-list items (to call appropriate `readPrec`). Is it even possible? In languages with reflection I can pass list of types (type - is usual value) and call type's methods, but how to solve similar in Haskell? === Best regards, Paul

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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 27 Nov '17

27 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 25 Nov '17

25 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 24 Nov '17

24 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 22 Nov '17

22 Nov '17

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

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Programming with Singletons
by Quentin Liu 19 Nov '17

19 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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