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Syntax error using where with guards.
by Manuel Vázquez Acosta 16 Sep '17

16 Sep '17

I was trying to implement a functional version of numerator/denominator simplifications of symbols. I came up with: simplify :: (Eq o, Ord o) => [o] -> [o] -> ([o], [o]) simplify [] bs = ([], bs) simplify ts [] = (ts, []) simplify (x:xs) (y:ys) | x == y = simplify xs ys | x < y = ([x] ++ (fst $ simplify xs (y:ys)), snd $ simplify xs (y:ys)) | otherwise = (fst r, [y] ++ snd r) where r = simplify (x:xs) ys simplify2 :: (Eq o, Ord o) => [o] -> [o] -> ([o], [o]) simplify2 a b = simplify (sort a) (sort b) -- sort not shown here In the second guard of the simplify function the expression 'simplify xs (y:ys)' is repeated. I would like to write instead: simplify (x:xs) (y:ys) | x == y = simplify xs ys | x < y = ([x] ++ fst r, snd r) where r = simplify xs (y:ys) | otherwise = (fst r, [y] ++ snd r) where r = simplify (x:xs) ys But doing so is parse error: $ ghc xoutil/dim/meta.hs [1 of 1] Compiling Meta ( xoutil/dim/meta.hs, xoutil/dim/meta.o ) xoutil/dim/meta.hs:19:3: parse error on input ‘|’ I'm using: ghc 7.10.3. Any ideas of why this is a syntax error? Best regards, Manuel.

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Square root algorithm
by mike h 15 Sep '17

15 Sep '17

Hi, To help me in learning Haskell I started blogging about some of the things I’ve looked at. One such topic was calculating square roots ‘by hand’ and then deriving a Haskell algorithm. I wrote about the well known technique here http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/ <http://gitcommit.co.uk/2017/08/25/the-root-of-the-problem-part-1/> and it it is really quite a simple method. The second part of the post will be an implementation in Haskell. I then tried implementing it and got something that works but really its not very pleasant to look at! And its something I don’t want to post! Some parts are fine but I think I locked myself into the notion that it had to be using State and really the end result is pretty poor. I know this i perhaps a ‘big ask’ but I’d really appreciate any suggestions, solutions, hints etc. I will of course give full attribution. I’ve created a gist of the code here https://gist.github.com/banditpig <https://gist.github.com/banditpig> Many Thanks Mike

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"Phantom" function?
by Baa 15 Sep '17

15 Sep '17

Hello, List. I'm trying to write function like this: type VerifyJson a = IO Bool isRight :: FromJSON a => FilePath -> VerifyJson a isRight testbed = do js <- readFile testbed return $ isJust (decode js) <<<<< ERROR IS HERE ! So I get error: 60 23 error error: • Could not deduce (FromJSON a0) arising from a use of ‘decode’ from the context: FromJSON a bound by the type signature for: isRight :: FromJSON a => FilePath -> VerifyJson a at /home/XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX/.stack-work/intero/intero7476qP4.hs:57:1-46 The type variable ‘a0’ is ambiguous These potential instances exist: instance FromJSON DotNetTime -- Defined in ‘aeson-1.0.2.1:Data.Aeson.Types.FromJSON’ instance FromJSON Value -- Defined in ‘aeson-1.0.2.1:Data.Aeson.Types.FromJSON’ instance (FromJSON a, FromJSON b) => FromJSON (Either a b) -- Defined in ‘aeson-1.0.2.1:Data.Aeson.Types.FromJSON’ ...plus 25 others ...plus 52 instances involving out-of-scope types (use -fprint-potential-instances to see them all) • In the first argument of ‘MB.isJust’, namely ‘(decode js)’ In the second argument of ‘($)’, namely ‘MB.isJust (decode js)’ In a stmt of a 'do' block: return $ MB.isJust (decode js) (intero) Is it possible to write such (phantom?;)) function? Sure, I can make workaround like: class FromJSON a => VerifyJson a where isRight :: FilePath -> IO Bool isRight testbed = do js <- readFile testbed return $ MB.isJust (decode js::Maybe a) but in this case I must declare instances for all verificating types (right?), like: instance VerifyJson MyType But idea was to minimize `isRight` users code... === Best regards, Paul

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understanding type constructors and value constructors
by Anwar Ludin 14 Sep '17

14 Sep '17

Hello everyone, I have just started studying Haskell and I am having a hard time understanding type and value constructors. So to create a new type, you write something like: data FinancialInstrument = Financial String Double deriving (Eq, Show) and then you can write: ibm = Financial "ibm" 150 OK all good. This initializes a FinancialInstrument. What I don't quite grasp is what is the purpose of Financial (the data/value constructor)? And from what I have read, you could have also written: data FinancialInstrument = FinancialInstrument String Double deriving (Eq, Show) To me the second expression is a lot closer to the typical OOP way of doing things (where the type name and constructor(s) have the same name). Why would someone prefer the first notation? Once a value has been constructed, how can I access its fields? Is there a way to create values using named parameters? Thanks!

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HSpec output option
by Baa 13 Sep '17

13 Sep '17

Hello, Dear List! I have tests: I'm using HSpec (and QuickCheck too). And I have tests like this: describe "Something" $ do it "something is correct" $ do ...blah-blah... it "any string is correct" $ property $ \s -> all (=='*') (Something s) -- it's for example only!!! so something like unit-test and property checks in one SomethingSpec.hs. I'm running them with this Makefile: .PHONY: test fast-test fast-test: stack exec runhaskell -- -isrc -itest test/Spec.hs test: stack test and in Spec.hs I have: {-# OPTIONS_GHC -F -pgmF hspec-discover #-} That's all. So, when I find failed test, I get a trace like this: ... Failures: test/SomethingSpec.hs:172: 1) BlahBlah.superFunc any string is correct: result Gave up after 48 tests ...etc... So, my question is: when QuichCheck runs my property test, it passes argument to property's lambda. And on 48th test attempt with some concreate argument value my check fails. How can I get detailed output from such test environment, to see what concreate arguments lead to failure? To see something (or similar/or more detailed even): Failed with arguments: s = "" Is it possible (I run them with stack and with runhaskell too) ? === Best regards, Paul

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Re: [Haskell-beginners] Beginners Digest, Vol 111, Issue 4
by Jack Brandt 04 Sep '17

04 Sep '17

Would a specialization of return work, as in f = return :: Maybe a -> IO (Maybe a)? On Mon, Sep 4, 2017 at 7:00 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. I have a -> f a. How to get m a -> f m a ? (Baa) > 2. Re: I have a -> f a. How to get m a -> f m a ? (Imants Cekusins) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Mon, 4 Sep 2017 15:06:34 +0300 > From: Baa <aquagnu(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] I have a -> f a. How to get m a -> f m a > ? > Message-ID: <20170904150634.230dd5ff@Pavel> > Content-Type: text/plain; charset=US-ASCII > > Hello List! > > I have function a -> IO a. How to get function: > > Maybe a -> IO (Maybe a) ? > > I found in Haskell mails archive such thing: > > class IFunctor f where > imap :: Idiom i => (s -> i t) -> f s -> i (f t) > > which looks similar, but I didn't find any helpfull instances of > `IFunctor` class in its package (and unfortunately I don't know what are > the indexed types: IMonad, IFunctor, etc). Sure, there is the primitive > solution like: > > myfunc :: a -> IO a > ... > f x = case x of Nothing -> return Nothing > Just x' -> Just <$> myfunc x' > > but more interesting is to know more standard and Haskelish solution > (like lift's, etc). May be I miss something very obvious.. > > === > Best regards, Paul > > > ------------------------------ > > Message: 2 > Date: Mon, 4 Sep 2017 15:14:42 +0300 > From: Imants Cekusins <imantc(a)gmail.com> > To: The Haskell-Beginners Mailing List - Discussion of primarily > beginner-level topics related to Haskell <beginners(a)haskell.org> > Subject: Re: [Haskell-beginners] I have a -> f a. How to get m a -> f > m a ? > Message-ID: > <CAP1qinaEhE6HqvS0O9-CfV9Vc2x8ZvfQ_SRFXQgVH- > yhapZZtg(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > > I have function a -> IO a. How to get function: > > Maybe a -> IO (Maybe a) ? > > Will mapM work: > > http://hackage.haskell.org/package/base-4.10.0.0/docs/ > Data-Traversable.html#v:mapM > > ? > > > > On 4 September 2017 at 15:06, Baa <aquagnu(a)gmail.com> wrote: > > > Hello List! > > > > I have function a -> IO a. How to get function: > > > > Maybe a -> IO (Maybe a) ? > > > > I found in Haskell mails archive such thing: > > > > class IFunctor f where > > imap :: Idiom i => (s -> i t) -> f s -> i (f t) > > > > which looks similar, but I didn't find any helpfull instances of > > `IFunctor` class in its package (and unfortunately I don't know what are > > the indexed types: IMonad, IFunctor, etc). Sure, there is the primitive > > solution like: > > > > myfunc :: a -> IO a > > ... > > f x = case x of Nothing -> return Nothing > > Just x' -> Just <$> myfunc x' > > > > but more interesting is to know more standard and Haskelish solution > > (like lift's, etc). May be I miss something very obvious.. > > > > === > > Best regards, Paul > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > >

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get rid of IO in [IO XXX]
by info＠maximka.de 04 Sep '17

04 Sep '17

What is the way to transform a list of [IO XXX] type to [XXX]? Thanks, Alexei

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I have a -> f a. How to get m a -> f m a ?
by Baa 04 Sep '17

04 Sep '17

Hello List! I have function a -> IO a. How to get function: Maybe a -> IO (Maybe a) ? I found in Haskell mails archive such thing: class IFunctor f where imap :: Idiom i => (s -> i t) -> f s -> i (f t) which looks similar, but I didn't find any helpfull instances of `IFunctor` class in its package (and unfortunately I don't know what are the indexed types: IMonad, IFunctor, etc). Sure, there is the primitive solution like: myfunc :: a -> IO a ... f x = case x of Nothing -> return Nothing Just x' -> Just <$> myfunc x' but more interesting is to know more standard and Haskelish solution (like lift's, etc). May be I miss something very obvious.. === Best regards, Paul

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Re: [Haskell-beginners] Restrict type in phantom data-type
by Baa 01 Sep '17

01 Sep '17

I pereputal emails. --- David, hello! 1. Is it the same/different as: data family Day a data Sunny data Rainy data instance Day Sunny = SunnyDay deriving Show data instance Day Rainy = RainyDay deriving Show ..and here you can not create `Day Int` object because no `Day Int` constructor (but you can create such constructor) ? Or in case with type families there is possibility to extend it to `Day Int` and in case with DayaKinds it's totally impossible? 2. I read somewhere (on forums) that restrictions on data types... I don't remember exactly, but something like they are not real restrictions or are related to old extension which is/will be deprecated. I'm not sure. Also, I'm not sure is it - in your example - restriction (constraint) or something else. Am I wrong? > This is maybe edging toward haskell-cafe territory, but you can > definitely do this in haskell. > > {-# LANGUAGE DataKinds, KindSignatures #-} > > data DayType = Sunny | Rainy > > data Day (a :: DayType) = Day > > > sunnyDay :: Day Sunny > sunnyDay = Day > > rainyDay :: Day Rainy > rainyDay = Day > > -- impossibleDay :: Day () > -- impossibleDay = Day > > On Fri, Sep 1, 2017 at 10:18 AM, Baa <aquagnu(a)gmail.com> wrote: > > Hello, List! > > > > For example, I have specialized (right nameis phantom?) type: > > > > data Day a = Day { ... no `a` here } > > data Sunny > > data Rainy > > > > joyToday :: Day Sunny -> IO () > > joyToday day = ... > > > > melancholyToday :: Day Rainy -> IO () > > melancholyToday day = ... > > > > And I can create (in spite of that it's phantom) some day: > > > > let day1 = Day {...} :: Day Sunny > > joyToday day1 > > > > but no problem to create `Day Int`, `Day Char`, etc which is > > pointless actually (sure "creator"-function can be exported from the > > module only, but I'm talking about type-level solution). > > > > I know that constraints (`... =>`) on data types are > > redundant/removed from the language. And I'm not sure how it's > > possible to restrict that parameter `a` (I know that it's possible > > to Java/C++/Perl6 (not sure), some other languages but how to add > > such restriction in Haskell? IMHO type families can help but I'm > > not sure how it will look (Sunny, Rainy are "nullary" type, so...). > > > > Is it possible for Haskell too? > > > > === > > Best regards, Paul > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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Restrict type in phantom data-type
by Baa 01 Sep '17

01 Sep '17

Hello, List! For example, I have specialized (right nameis phantom?) type: data Day a = Day { ... no `a` here } data Sunny data Rainy joyToday :: Day Sunny -> IO () joyToday day = ... melancholyToday :: Day Rainy -> IO () melancholyToday day = ... And I can create (in spite of that it's phantom) some day: let day1 = Day {...} :: Day Sunny joyToday day1 but no problem to create `Day Int`, `Day Char`, etc which is pointless actually (sure "creator"-function can be exported from the module only, but I'm talking about type-level solution). I know that constraints (`... =>`) on data types are redundant/removed from the language. And I'm not sure how it's possible to restrict that parameter `a` (I know that it's possible to Java/C++/Perl6 (not sure), some other languages but how to add such restriction in Haskell? IMHO type families can help but I'm not sure how it will look (Sunny, Rainy are "nullary" type, so...). Is it possible for Haskell too? === Best regards, Paul

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