- Beginners - Haskell.org

Re: [Haskell-beginners] Beginners Digest, Vol 122, Issue 6
by Anthony Lee 17 Aug '18

17 Aug '18

The source code is here: https://github.com/ekmett/bound/ On Fri, Aug 17, 2018 at 8:41 AM <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Bound library questions (Anthony Lee) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Fri, 17 Aug 2018 06:46:52 -0400 > From: Anthony Lee <anthonynlee(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Bound library questions > Message-ID: > <CA+pBo5HVraf_8jbs15bR6QAy1S7=z_u-3KBwT= > c3sqkAgc0m5Q(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > In Scope.hs there are some functions I feel difficult to understand, > Why fmap/foldmap/traverse is applied three times? > > instance Functor f => Functor (Scope b f) where > fmap f (Scope a) = Scope (fmap (fmap (fmap f)) a) > {-# INLINE fmap #-} > > -- | @'toList'@ is provides a list (with duplicates) of the free variables > instance Foldable f => Foldable (Scope b f) where > foldMap f (Scope a) = foldMap (foldMap (foldMap f)) a > {-# INLINE foldMap #-} > > instance Traversable f => Traversable (Scope b f) where > traverse f (Scope a) = Scope <$> traverse (traverse (traverse f)) a > {-# INLINE traverse #-} >

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Exercise 7.2a
by trent shipley 15 Aug '18

15 Aug '18

I am trying to solve Hutton 2016 exercise 7.2a. My attempt gets the same error message as the one below. The problem with the error message below is that it is for the book's solved example, so I have no idea how to make it work. I suspect I have GHCi configured wrong, or an old version that is buggy on Mac, or such. Help is appreciated. Also, at this point I am having trouble reading these function declarations. all :: (a -> Bool) -> [Bool] -> Bool all b ls = and (map b ls) My try: all :: -- name of function defined (a -> -- maps to "all" Bool -- maps to boolean function b ) -> [Bool] -- maps to input ls -> Bool -- maps to return type of "all" all b ls = and (map b ls) OR (right?) all :: -- name of function defined (a -> Bool) function of type a returns bool -> [Bool] -- maps to input ls -> Bool -- maps to return type of "all" all b ls = and (map b ls) ------ Book example all :: (a -- maps to "all" -> Bool -- input function type ) -> [Bool] -- implied list of things mapped over -> Bool -- returned all p = and . map p OR (right?) all :: -- name of defined function (a -> Bool) -- input function of type a returns type Bool -> [Bool] -- implied list to work on -> Bool -- return type all p = and . map p ============================ -- CODE {-- 2. Without looking at the definitions from the standard prelude, define the following higher-order library functions on lists. a.Decide if all elements of a list satisfy a predicate: all :: (a -> Bool) -> [Bool] -> Bool b.Decide if any element of a list satisfies a predicate: any :: (a -> Bool) -> [Bool] -> Bool c.Select elements from a list while they satisfy a predicate: takeWhile :: (a -> Bool) -> [a] -> [a] d.Remove elements from a list while they satisfy a predicate: dropWhile :: (a -> Bool) -> [a] -> [a] Note: in the prelude the first two of these functions are generic functions rather than being specific to the type of lists. Hutton, Graham. Programming in Haskell (Kindle Locations 2806-2819). Cambridge University Press. Kindle Edition. --} import Prelude hiding (all) {-- My attempt. Presumably bad. -- I swear GHCi was happy with it yesterday. all :: (a -> Bool) -> [Bool] -> Bool all b ls = and (map b ls) --} -- from book all :: (a -> Bool) -> [Bool] -> Bool all p = and . map p {-- >From GHCi on Mac Prelude> :r [1 of 1] Compiling Main ( ex7_2a.hs, interpreted ) ex7_2a.hs:30:9: error: • Couldn't match type ‘a’ with ‘Bool’ ‘a’ is a rigid type variable bound by the type signature for: all :: forall a. (a -> Bool) -> [Bool] -> Bool at ex7_2a.hs:29:1-36 Expected type: [Bool] -> Bool Actual type: [a] -> Bool • In the expression: and . map p In an equation for ‘all’: all p = and . map p • Relevant bindings include p :: a -> Bool (bound at ex7_2a.hs:30:5) all :: (a -> Bool) -> [Bool] -> Bool (bound at ex7_2a.hs:30:1) | 30 | all p = and . map p | ^^^^^^^^^^^ Failed, no modules loaded. --}

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How to write the FromJSON instance for CUid
by PICCA Frederic-Emmanuel 10 Aug '18

10 Aug '18

Hello, I try to write the instance for the CUid[1] type But I do not know what to do. instance FromJSON CUid where parseJSON = ... {-# INLINE parseJSON #-} Thansk for your help [1] https://hackage.haskell.org/package/base-4.11.1.0/docs/System-Posix-Types.h…

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Re: [Haskell-beginners] understanding function signature alignement
by simkestuff＠gmail.com 10 Aug '18

10 Aug '18

> > Thanks, it is still a bit fuzzy to me ... I understand what you did but what confuses me is that when i look at function with signature like f :: Monad m => c -> m d I always think that return type is somehow restricted in comparison to input because it demands that output type is wraped inside something (monad in this case). For such signature to fit id signature (a -> a) , c type shoud be also wraped inside monad but it is not case here... Anyhow, I still have to figure it out thanks

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understanding function signature alignement
by simkestuff＠gmail.com 09 Aug '18

09 Aug '18

Hello, I'm going through Haskellbook and while doing some exercises I'm stack with trying to explain myself how types matches in this examples: meh :: Monad m => [a] -> (a -> m b) -> m [b] meh [] _ = pure [] meh (x:xs) f = (:) <$> f x <*> meh xs f flipType :: (Monad m) => [m a] -> m [a] flipType xs = meh xs id What puzzles me is how id function which is of type (a->a) can fit here where meh function is requesting function of type (a->m b)? The function (a -> m b) is function that says give me anything and I will give back anything wrapped up into some monad structure; id ( a -> a) on other hand says give me anything and I will return it back to you. So, to me, the first function is somehow more restricted than id function because it puts limitation on what output of that function can be and I'm struggling to understand how id function can fit here. I hope someone can help me how to reason about these functions here? thanks

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Best way to trig an IO action on a condition.
by PICCA Frederic-Emmanuel 29 Jul '18

29 Jul '18

Hello, Now that I know how to fork a process the right way :) I would like to trig this process when a list of file exist on the system. I can not use inotify since my files are on an nfs sytem. So I would like your advices in order to create a Condition which wait for the presence of these files and when this condition is fullliled, trigg the action. Thanks for your help. Frederic.

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Where is the accumulator in the expression foldr (<=<) return (replicate x oveKnight)? -- from LYAH example
by Olumide 27 Jul '18

27 Jul '18

Dear List, Chapter 13 of LYAH (http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions) has the following code block import Data.List inMany :: Int -> KnightPos -> [KnightPos] inMany x start = return start >>= foldr (<=<) return (replicate x oveKnight) What I'd like to know is where the accumulator of foldr is in this example. Regards, - Olumide

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forkProcess behaviour
by PICCA Frederic-Emmanuel 25 Jul '18

25 Jul '18

Hello, I try to write a service which execute an IO using forkProcess This service contain a web server whcih avwait for job published by users. for each job, I create a forkProcess of this job. I need to fork the process in order to change the uid and gid of the process for each of the job dependencing on who request the job. I have a least two questions 1) is this forkProcess a fork of all the current process, or is it just an executin of the IO in another process id ? I ask this because sometime one of my job hang and a process keep running. Since the original process is binded to a port, it is not possible to restart the server, saying that the port is already in use. 2 ) How can I catch the exceptions thrown from the child process in order to process them in the parent process. thanks for your help answering these questions. Frederic

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The type class Read
by mrx 13 Jul '18

13 Jul '18

Hi, I'm trying to make sense of the type class `Read`. It feels to be like a function performing casting to me, similar to what you have in C and Java. The weird thing is that the casting doesn't have to make sense, type-wise. So `read "5" :: Int` produce the integer 5 which make sense but `read "(3,'a')" :: Int` doesn't make sense to me but ghci accepts it! This is seeing Read as a function, as it's a type class i suspect I'm missing something. Quite possibly quite a lot... Why is the type class called `Read`? What am I missing above? // Patrik Iselind

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Two small questions from the section "Making a safe RPN calculator" in LYAH
by Olumide 11 Jul '18

11 Jul '18

Dear List, Question 1: In the section "Making a safe RPN calculator" of LYAH (Chapter 14 -- http://learnyouahaskell.com/for-a-few-monads-more#useful-monadic-functions) there is an expression: read numberString:xs, take from the last line of the function foldingFunction :: [Double] -> String -> [Double] foldingFunction (x:y:ys) "*" = (x * y):ys foldingFunction (x:y:ys) "+" = (x + y):ys foldingFunction (x:y:ys) "-" = (y - x):ys foldingFunction xs numberString = read numberString:xs My first question is why is the read function called before the cons operator? Question 2: The same section of the book also introduces the reads function which is used to implement the readMaybe function and a refactored foldingFunction readMaybe :: (Read a) => String -> Maybe a readMaybe st = case reads st of [(x,"")] -> Just x _ -> Nothing foldingFunction :: [Double] -> String -> Maybe [Double] foldingFunction (x:y:ys) "*" = return ((x * y):ys) foldingFunction (x:y:ys) "+" = return ((x + y):ys) foldingFunction (x:y:ys) "-" = return ((y - x):ys) foldingFunction xs numberString = liftM (:xs) (readMaybe numberString) I'd like to know why the foldingFunction returns Nothing in the following example: ghci> foldingFunction [] "1 wawawawa" Nothing Considering that reads "1 wawawawa" does not return Nothing, as follows ghci> reads "1 wawawawa" :: [(Double,String)] [(1.0," wawawawa")] Regards, - Olumide

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