December 2015 - Beginners

Pattern matching is actually about matching constructors??
by Olumide 22 Dec '15

22 Dec '15

Hello, On chapter 7 of LYH there's an example of a user-defined operator .++ infixr 5 .++ (.++) :: List a -> List a -> List a Empty .++ ys = ys (x :-: xs) .++ ys = x :-: (xs .++ ys) which is used as follows let a = 3 :-: 4 :-: 5 :-: Empty let b = 6 :-: 7 :-: Empty a .++ b (:-:) 3 ((:-:) 4 ((:-:) 5 ((:-:) 6 ((:-:) 7 Empty)))) Following this the text reads: "Notice how we pattern matched on (x :-: xs). That works because pattern matching is actually about matching constructors. We can match on :-: because it is a constructor for our own list type ..." Source: http://learnyouahaskell.com/making-our-own-types-and-typeclasses#recursive-… Is the operator :-: a constructor? I'm confused because the definition of :-: is not prefixed by the data keyword? - Olumide

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Mixing pattern matching and guards
by martin 22 Dec '15

22 Dec '15

Hello all, I recently wrote this piece of code import qualified Data.Set as S [...] purge :: (Eq cp, Eq ip) => Product cp ip -> Product cp ip purge (Punion ps) | ps' == S.empty = Pempty where ps' = S.filter (/= Pempty) ps purge x = x and I thought I had missed the "otherwise" case in the guard and I was prepared to see a "non exhaustive ..." error, but to my amazement it works *Main> purge $ Punion $ S.fromList [Packed 1 Pempty] Punion (fromList [Packed 1 Pempty]) As this is a Punion, it should match the first pattern, but not the guard. It seems to fall right through to the second pattern. Is this the way it works?

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Re: [Haskell-beginners] How to nest arbitrary things
by Imants Cekusins 21 Dec '15

21 Dec '15

> how to write an "unpack" function is this what Lenses are all about? https://www.fpcomplete.com/school/to-infinity-and-beyond/pick-of-the-week/a… tutorial I stumbled across Lenses only recently so can not write a working example yet. if Lenses are relevant to this topic, could someone kindly post an appropriate (Truck Parcel Can) datatype and an "unpack" with Lenses example?

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How to nest arbitrary things
by martin 21 Dec '15

21 Dec '15

Hello all, I was trying to model things which can contain other things. This is easy as long as containers and the contained items all can be described in the same fashion. However when I want to model - say - trucks containing boxes containing parcels containing cans and trucks, boxes, parcels and cans are not of the same type, then this nested thing will become a type in its own right and it will be of a different type than trucks containing cans (which are not wrappen in parcels ...) As long as I can squeeze trucks, boxes ... into one type, possibly by using a "container_type" attribute, there is no problem. Is this the only way to do this? Is there an idiom?

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Problem reading sound data from mic with TChan
by Martin Vlk 21 Dec '15

21 Dec '15

Hi, I am working on a little toy project related to controlling graphics display based on data from the computer microphone. I am basing the whole thing on concepts from game programming, so I have a main loop which reads inputs, updates the world state and generates outputs. As part of inputs there is microphone. I read data from it using the pulse-simple library. The "simpleRead" function blocks if there is not enough data available so I can't use it directly in the main loop or I risk delays. So I figured I'll use a separate thread to read from the mic and write data into a TChan. The main loop in separate thread then can read from the TChan as needed and test for availability of data to avoid delaying the main loop. Here is my current code: http://lpaste.net/147522 The data is written into TChan in the "handleMic" function and read from the TChan on line 85. The problem I have is that the TChan never seems to contain any data when I read from it and that confuses me. Why? Does anyone see where is my problem? Many Thanks Martin

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Questions About Rose Trees Analysis
by Lorenzo Isella 20 Dec '15

20 Dec '15

Dear All, I am trying to learn Haskell (for the second time), but I have to admit I am still struggling with relative simple stuff, so I ask if anybody can give me a hand. I am dealing with the issues of extracting info out of some rose trees and probably I am also having some troubles with the notation. The rose tree is defined with the following notation data Rose a = a :> [Rose a] deriving (Eq, Show) and the root can be detected simply as root (a :> rs) = a I would like to have the expression (with the ":>" notation for the Node) of the function to find the children of the root. Example of expected behavior of this function children children (1 :> [2 :> [], 3 :> []]) = [2 :> [], 3 :> []] On top of that, I am trying to get the functions to have the functions size :: Rose a -> Int leaves :: Rose a -> Int that count the number of nodes in a rose tree, respectively the number of leaves (nodes without any children). For the children function, I have tried stuff like children (a :> rs) = rs quite unsuccessfully. Any suggestion is appreciated. Cheers Lorenzo

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Learning Haskell: Algebraic Data Types
by Manuel M T Chakravarty 18 Dec '15

18 Dec '15

The latest chapter of our ”Learning Haskell” tutorial takes a first stab at algebraic data types — there will be a second chapter continuing the topic sometime soon: http://blog.haskellformac.com/blog/algebraic-data-types-demystified Happy Coding! Manuel

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Extend instance for List
by Graham Gill 16 Dec '15

16 Dec '15

The NICTA course <https://github.com/NICTA/course> includes exercises on the type class Extend, in Course/Extend.hs. Extend is a superclass of Comonad. Here's the class definition: > -- | All instances of the `Extend` type-class must satisfy one law. > This law > -- is not checked by the compiler. This law is given as: > -- > -- * The law of associativity > -- `∀f g. (f <<=) . (g <<=) ≅ (<<=) (f . (g <<=))` > class Functor f => Extend f where > -- Pronounced, extend. > (<<=) :: > (f a -> b) > -> f a > -> f b > > infixr 1 <<= Could someone please motivate the Extend instance for List? (Its implementation is left as an exercise. In the course, type List a is isomorphic to [a].) Some of the tests (<<=) is expected to pass are shown, and make clear what ought to happen. > -- | Implement the @Extend@ instance for @List@. > -- > -- >>> length <<= ('a' :. 'b' :. 'c' :. Nil) > -- [3,2,1] > -- > -- >>> id <<= (1 :. 2 :. 3 :. 4 :. Nil) > -- [[1,2,3,4],[2,3,4],[3,4],[4]] > -- > -- >>> reverse <<= ((1 :. 2 :. 3 :. Nil) :. (4 :. 5 :. 6 :. Nil) :. Nil) > -- [[[4,5,6],[1,2,3]],[[4,5,6]]] The following (wrong, according to the tests) Extend instance for List nevertheless obeys the types and obeys the Extend law of associativity. > instance Extend List where > (<<=) :: > (List a -> b) > -> List a > -> List b > (<<=) f = (:. Nil) . f (:. Nil) is analogous to (: []), create a singleton list. I can't find a good reference on the Extend type class to convince me why the correct Extend instance for List in the course is the desirable one. (I'm not saying my version is desirable, in fact it seems fairly useless, but it works.) Graham

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Monad and bind operator interpretation
by Raja 15 Dec '15

15 Dec '15

The type signature of bind (>>=) is as follows: (>>=) :: m a -> (a -> m b) -> m b One interpretation of this could be as follows: bind takes two parameters (m a & f) and returns m b (the same type returned by f) So extending this interpretation - can I swap the two parameters (?) Now my new hypothetical interpretation becomes: (>>=) :: (a -> m b) -> m a -> m b If i further add parens: (>>=) (a -> m b) -> (m a -> m b) This allows me to slightly tweak my interpretation: bind takes one param f (of type a -> m b) and returns another param f (of type m a -> m b) This feels like a more intuitive way to think about Monads - am I on the right track? (not that I want to switch the params permanently - just trying to get a feel for monads)

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IO a, IO b
by Imants Cekusins 15 Dec '15

15 Dec '15

Hello all, This looks like another potential 1+ day problem for me. Could you please help? Problem: a few (... -> IO a) functions are called as part of the same "do" block of container function these functions may return various IO a: "a" can be of different types I need the return values to proceed i.e. would call val <- (... -> IO a) container function returns IO b specifically, let's consider 2 functions called inside do block: ... -> IO Bool ... -> IO (Either String a) container function returns IO (Either String a) Questions: 1) what is a beginner-friendly way of doing this? 2) how would you do this? links, ideas or snippets will all be appreciated.

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