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Re: [Haskell-beginners] Unpacking tuples
by Frerich Raabe 04 Dec '17

04 Dec '17

On 2017-12-04 20:05, mrx wrote: > Hi, > > If I have a function that produce tuples with three members and I want those > members as parameters for a function that takes three parameters. > How would I unpack that tuple? > > It seems that curry does the trick for tuples with two members. > > How do I do this when there are more than two members? There are functions like 'curry3' ( http://hackage.haskell.org/package/utility-ht-0.0.14/docs/Data-Tuple-HT.htm… ) but I think if you don't need this very often, it might be easiest to just go for let (a,b,c) = f x in g a b c -- Frerich Raabe - raabe(a)froglogic.com www.froglogic.com - Multi-Platform GUI Testing

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Unpacking tuples
by mrx 04 Dec '17

04 Dec '17

Hi, If I have a function that produce tuples with three members and I want those members as parameters for a function that takes three parameters. How would I unpack that tuple? It seems that curry does the trick for tuples with two members. How do I do this when there are more than two members? Patrik

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"Shortcut" for f (g x) (g x)
by Baa 04 Dec '17

04 Dec '17

Hello, All! Does Haskell have some "short-form" for such call: `f (g x) (g x)`, for example: compare (snd x) (snd x) Looks like combinatory logic or `ap` but not exactly... === Best regards, Paul

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Overlapping instances problem
by Baa 01 Dec '17

01 Dec '17

Hello, List! I got error: Duplicate instance declarations: instance [overlap ok] EnumTag a => Read a -- Defined at /XXX/intero/intero2932Xpa-TEMP.hs:110:27 instance [overlap ok] StrTag a => Read a -- Defined at /XXX/intero/intero2932Xpa-TEMP.hs:121:27 (intero) For this code: class (Show a, Enum a) => EnumTag a where anyEnum :: a instance {-# OVERLAPS #-} EnumTag a => Read a where readPrec = RP.lift P.skipSpaces >> expectEnum instance {-# OVERLAPS #-} EnumTag a => Eq a where a == b | a == anyEnum || b == anyEnum = True | otherwise = fromEnum a == fromEnum b class StrTag a where anyStr :: a tagPrefix :: a -> String -- ^ should be constant toStr :: String -> a instance {-# OVERLAPS #-} StrTag a => Read a where readPrec = parens $ do RP.lift P.skipSpaces (RP.lift $ expectShown anyStr) <++ RP.lift g where g = do Just s@(_:_) <- L.stripPrefix tagPrefix <$> expectTag return $ toStr s Why does it happen? `Read a` in 1st instance is valid only when a is `EnumTag`, in 2nd one - is valid only when a is `StrTag`. How can I fix this error and to create "default" instances for `EnumTag` and to `StrTag`, so client code will "inherit" those functionality (`Read`) simple, only with instantiation of `EnumTag` or `StrTag` ? Sure, if I comment `instance ... StrTag a` then all work fine, but I need 2 specialized `Read`s (and `Eq`s too :) === Best regards, Paul

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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 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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