- Beginners - Haskell.org

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by Christopher Ledbetter 27 Apr '20

27 Apr '20

Unsubscribe Sent from my iPad > On Apr 26, 2020, at 8:51 AM, Ken Overton <ken.overton(a)gmail.com> wrote: > > > Hello all, > > I recently came across this function which made me realize I don't understand list comprehensions well. I hope someone can help me understand them better by understanding this example better. The function takes a list of Eq and returns the list of unique elements from it: > > unique :: Eq a => [a] -> [a] > unique xs = [x | (x,y) <- zip xs [0..], x `notElem` (take y xs)] > > It's using a list comprehension with multiple 'generators' (hope I have the term correctly). My understanding of multiple generators in a list comprehension is that they refine the results of the previous generator. > > So the first generator should produce [(Eq,Int)] as input to the second generator? And the second generator should produce [Bool]? > > My understanding must be wrong though; how do we end up with just the items where the second generator produced True? > > Thanks, > > > -- > Ken Overton > (917) 863-3937 > ken.overton(a)gmail.com > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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List comprehensions with multiple generators
by Ken Overton 27 Apr '20

27 Apr '20

Hello all, I recently came across this function which made me realize I don't understand list comprehensions well. I hope someone can help me understand them better by understanding this example better. The function takes a list of Eq and returns the list of unique elements from it: unique :: Eq a => [a] -> [a] unique xs = [x | (x,y) <- zip xs [0..], x `notElem` (take y xs)] It's using a list comprehension with multiple 'generators' (hope I have the term correctly). My understanding of multiple generators in a list comprehension is that they refine the results of the previous generator. So the first generator should produce [(Eq,Int)] as input to the second generator? And the second generator should produce [Bool]? My understanding must be wrong though; how do we end up with just the items where the second generator produced True? Thanks, -- Ken Overton (917) 863-3937 ken.overton(a)gmail.com

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instance 2
by Alexander Chen 23 Apr '20

23 Apr '20

hi, data StringOrInt = TisanInt Int | TisAString String instance Eq (StringOrInt) where (==) (TisanInt v) (TisanInt v') = v == v' (==) (TisAString s) (TisAString s') = s == s' prelude> Instance_testen.hs:4:37: error: Ambiguous occurrence ‘==’ It could refer to either ‘Prelude.==’, imported from ‘Prelude’ at Instance_testen.hs:1:1 (and originally defined in ‘GHC.Classes’) or ‘Main.==’, defined at Instance_testen.hs:4:1 | 4 | (==) (TisanInt v) (TisanInt v') = v == v' | ^^ Instance_testen.hs:5:41: error: Ambiguous occurrence ‘==’ It could refer to either ‘Prelude.==’, imported from ‘Prelude’ at Instance_testen.hs:1:1 (and originally defined in ‘GHC.Classes’) or ‘Main.==’, defined at Instance_testen.hs:4:1 | 5 | (==) (TisAString s) (TisAString s') = s == s' | ^^ [1 of 1] Compiling Main ( Instance_testen.hs, interpreted ) Failed, no modules loaded. what gives?

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instance
by Alexander Chen 23 Apr '20

23 Apr '20

Hi, data TisanInteger = Tisan Integer instance Eq (Tisan Integer) where (Tisan v) == (Tisan v') = v == v' prelude> Instance_testen.hs:7:14: error: Not in scope: type constructor or class ‘Tisan’ A data constructor of that name is in scope; did you mean DataKinds? | 7 | instance Eq (Tisan Integer) where (Tisan v) == (Tisan v') = v == v' | ^^^^^ [1 of 1] Compiling Main ( Instance_testen.hs, interpreted ) Failed, no modules loaded. what am i doing wrong? best,

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[ANN] Haskell Beginners WT.Social Community: "Haskell: The Haskell-Beginners Community"
by Benjamin L. Russell 19 Apr '20

19 Apr '20

Twenty-eight days ago (on March 23, 2020 JST), I created a new Haskell beginners WT.Social community, "Haskell: The Haskell-Beginners Community" (see https://pashpost.com/group/haskell-the-haskell-beginners-community) and migrated a substantial portion of the posts originally from my former corresponding pashPOST (since renamed to "theTUNDRA") community of the same name thereto (sansposts with missing links or irrelevant content). Feel free to join and contribute! -- Benjamin L. Russell -- Benjamin L. Russell / DekuDekuplex at Yahoo dot com http://dekude kuplex.wordpress.com/ Computer Science Document Translator/Editor "Furuike ya, kawazu tobikomu mizu no oto." -- Matsuo Basho^

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Why the compiler does not commit to a type class?
by Alexander Chen 18 Apr '20

18 Apr '20

hi, Prelude> u = undefined Prelude> k :: (Ord a, Num b) => a -> b -> a; k = u Prelude> :t k 1 2 k 1 2 :: (Ord a, Num a) => a Prelude> :t k (1 :: Integer) 2 k (1 :: Integer) 2 :: Integer In the first version the compiler is not really pushed so it simply gives back the type class constriction it has to adhere to. While in the second it is pushed to give an more specific answer and thus the :: Integer. Question: Is this train of thought correct?

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'/' instead use 'DIV'. why?
by Alexander Chen 18 Apr '20

18 Apr '20

Hi, Prelude> :t (/) (/) :: Fractional a => a -> a -> a Prelude> :t div div :: Integral a => a -> a -> a Prelude> 6 / length [23,34,45] error Prelude> 6 / 3 2.0 Could somebody explain to me why this is? thanks, Alexander Chen

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Factor out parsec types
by Max Gautier 25 Mar '20

25 Mar '20

I'm writing a parser for polynomial equation (ex : 5 x ^2 + 3 x ^ 1 = 3 x ^ 2). I'm using parsec to decompose my parser into smaller parts, and it works, but all my parser functions have the type : parser :: Stream s m Char => ParsecT s u m <Type to Parse> Is there a way to factor out that boilerplate, in such a way that I could simply write a type like parser :: Parser <Type to parse> ? The farther I've been is : type Parser s = ParsecT s () Identity type St = Stream s Identity Char (using ConstraintKinds) parser :: St s => Parser s <Type to Parse> But that still feels tedious to write every time... I've tried to search the wiki and the internet in general, but did not find an answer to that particularly. Thxs a lot if you have any hindsights. Max Gautier

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Re: [Haskell-beginners] curious behavior of Haskell wiki
by Doug McIlroy 24 Mar '20

24 Mar '20

Yes, you repeated what I said. That search finds nothing. But it is wrong. A search for "bitwise" finds https://wiki.haskell.org/Ord_instance The context of "bitwise" on that page is "bitwise AND". My question remains. Date: Tue, 24 Mar 2020 01:33:48 +0100 From: "Henk-Jan van Tuyl" <b.vantuyl1(a)upcmail.nl> If you enter "bitwise AND" with the quotes, the wiki searches for these words next to each other; there is no page with such an occurrence. Regards, Henk-Jan van Tuyl On Sat, 21 Mar 2020 15:18:07 +0100, Doug McIlroy <doug(a)cs.dartmouth.edu> wrote: > Why does a search for "bitwise" in the Haskell wiki find > an instance of "bitwise AND", whereas a search for > "bitwise AND" or "bitwise and" (in quotes) finds nothing? > > Is this a bug or a (mis)feature? > > Doug McIlroy

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Different versions of ghci REPL in Emacs?
by Lawrence Bottorff 22 Mar '20

22 Mar '20

I'm using Emacs (26.3) and org-mode (9.3.6) and I seem to have two different versions of ghci going, depending on whether I'm using haskell-mode alone with a *.hs file, or whether I am evaluating an org-mode babel code block. If I open, say, test.hs in Emacs, haskell-mode starts up. And if I choose <Start interpreter> from the menu, I get this "lambda prompt" REPL The next big Haskell project is about to start! If I break, you can: 1. Restart: M-x haskell-process-restart 2. Configure logging: C-h v haskell-process-log (useful for debugging) 3. General config: M-x customize-mode 4. Hide these tips: C-h v haskell-process-show-debug-tips λ> But if I call M-run-haskell, I get this "Prelude prompt" REPL Configuring GHCi with the following packages: GHCi, version 8.6.5: http://www.haskell.org/ghc/ :? for help Loaded GHCi configuration from /home/lawrence/.ghc/ghci.conf Prelude> Prelude> and this babel bit knows and interacts with it #+BEGIN_SRC haskell :results output :set +m doubleMe x = x + x doubleUs x y = doubleMe x + doubleMe y #+END_SRC but then trying to run this code block with lambda prompt REPL, I get the error executing Haskell code block... org-babel-execute:haskell: Buffer *haskell* does not exist or has no process even though the lambda REPL is called *haskell*. Why am I getting two separate REPLs? BTW, which IDE do Haskell big boys and girls use? LB

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