January 2017 - Beginners

What is wrong with my types signatures.
by PICCA Frederic-Emmanuel 23 Jan '17

23 Jan '17

Hello here a snipset code class Shape sh => FrameND t sh where shapeND :: t -> MaybeT IO sh rowND :: t -> sh -> MaybeT IO (XrdMeshFrame sh) I have an instance of FrameND then I created a function which use this class framesND :: (Shape sh, FrameND a sh) => Pipe a (XrdMeshFrame sh) IO () framesND = do d <- await sh' <- lift $ runMaybeT $ shapeND d let n = size sh' forM_ [0..n-1] (\i' -> do f <- lift $ runMaybeT $ rowND d (fromIndex sh' i') when (isJust f) (yield (fromJust f))) But when I compile it; I get this error message src/Hkl/XRD.hs:613:55: Could not deduce (sh ~ Maybe a0) from the context (Shape sh, FrameND a sh) bound by the type signature for framesND :: (Shape sh, FrameND a sh) => Pipe a (XrdMeshFrame sh) IO () at src/Hkl/XRD.hs:606:13-70 `sh' is a rigid type variable bound by the type signature for framesND :: (Shape sh, FrameND a sh) => Pipe a (XrdMeshFrame sh) IO () at src/Hkl/XRD.hs:606:13 Expected type: Maybe (XrdMeshFrame sh) Actual type: Maybe (XrdMeshFrame (Maybe a0)) In the first argument of `fromJust', namely `f' In the first argument of `yield', namely `(fromJust f)' In the second argument of `when', namely `(yield (fromJust f))' So I do not undestand why the typse system guess Maybe a0 instead of sh in f If I read this f <- lift $ runMaybeT $ rowND d (fromIndex sh' i') runMaybeT return a (Maybe (XrdMeshFrame sh)) and then I lift it into the Pipe. So What is wrong ? thanks Frederic

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how to skip pattern match error when applying a mapM_
by PICCA Frederic-Emmanuel 18 Jan '17

18 Jan '17

Sorry I forgot to put a subject ________________________________________ De : Beginners [beginners-bounces(a)haskell.org] de la part de PICCA Frederic-Emmanuel Envoyé : mardi 17 janvier 2017 11:15 À : The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell Objet : [Haskell-beginners] (no subject) Hello, Here a reduction of my problem values :: IO [IO (Maybe Int)] values = do let v = [Just 1, Just 2, Just 3, Nothing, Just 5, Nothing, Just 7] :: [Maybe Int] return $ map return v main :: IO () main = do vs <- values nvs <- mapM_ go vs print nvs where go :: IO (Maybe Int) -> IO Int go v' = do Just v <- v' return v when I run this script, I get a runtime error picca@diffabs6:~/tmp$ runhaskell test.hs test.hs: user error (Pattern match failure in do expression at test.hs:13:10-15) What I want is a go method which skip silently the (IO Nothing) values. so when used in the mapM_ it return only the values which are returned by the IO (Maybe Int) (stored in the values) Thanks for your help Frédéric Indeed _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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Feedback needed - CLI app + guide
by Carlo Matteo Scalzo 16 Jan '17

16 Jan '17

Hi all, I'm new to the community here, so please do let me know if I'm doing something wrong :-) I've been learning Haskell over the last year or so, reading books etc., but I couldn't find a simple, real world example, so I decided to to write a simple command-line application in Haskell. The application and the code are free, and I've also written a short guide that uses the application as a use case to talk about Haskell and its features. Can you guys please download the guide (for free, of course) , and let me know if you found it useful? You can find it here: https://fifohaskell.com Please let me have your feedback (my personal email is on the website, if you don't want to reply here) - and thank you very much for your help! Carlo

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safe versions of pred and succ?
by Graham Gill 16 Jan '17

16 Jan '17

Suppose I have a function f :: (Enum a) => a -> ... f x ... = ...pred x ... If a is also an instance of Bounded, then I get a runtime error if pred minBound is evaluated during evaluation of f. If I try to protect the use of pred x with a check using minBound, then I have to add a Bounded constraint on type a, which would mean, for example, that I can no longer use f with type Integer. Do I need two different versions of f, one for Bounded a and one for non-Bounded a? Is there a more elegant way to take care of this problem? I don't know much about all of the type magic available in GHC. For example, from another list message, groupConsecutive :: (Enum a,Eq a) => [a] -> [[a]] groupConsecutive = foldr go [] where go x ls@(hd@(y:_):yss) | x == y || x == pred y = (x:hd):yss | otherwise = [x]:ls go x [] = [[x]] > groupConsecutive [1,2,3,7,8,10,11,12] [[1,2,3],[7,8],[10,11,12]] > groupConsecutive ([1,0,1,2]::[Word]) *** Exception: Enum.pred{Word}: tried to take `pred' of minBound In the first go case, if type a is also Bounded, y == minBound and x /= y, then we know already that x /= pred y, we want the guard to fail and that we pass to the otherwise guard to start a new sublist. But how to implement it without making the function unavailable for un-Bounded types? Graham

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Re: [Haskell-beginners] Beginners Digest, Vol 103, Issue 8
by Saqib Shamsi 16 Jan '17

16 Jan '17

Hello Francesco, Sorry for the late reply. Yes, your solution is elegant and does what I wanted. This is what I did -- zipped is the list of tuples formed as you mentioned λ> groupBy (\x y -> (snd x) == (snd y)) zipped [[(1,0),(2,0),(3,0)],[(7,3),(8,3)],[(10,4),(11,4),(12,4)]] λ> [[fst tpl | tpl <- x] | x <- it] [[1,2,3],[7,8],[10,11,12]] This works perfectly fine. Thank you. Regards, Saqib Shamsi On 14 January 2017 at 00:27, <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. Better Code (Saqib Shamsi) > 2. Re: Better Code (Francesco Ariis) > 3. Re: Better Code (Joel Neely) > 4. Re: Better Code (Joel Neely) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Fri, 13 Jan 2017 21:35:54 +0530 > From: Saqib Shamsi <shamsi.saqib(a)gmail.com> > To: beginners(a)haskell.org > Subject: [Haskell-beginners] Better Code > Message-ID: > <CAKpYsCGKeWoHttqpZAF7jAkhe++PqLSfwZA820tA=Je8hFiu8g@mail. > gmail.com> > Content-Type: text/plain; charset="utf-8" > > Hi, > > The problem that I wish to solve is to divide a (sored) list of integers > into sublists such that each sublist contains numbers in consecutive > sequence. > > For example, > *Input:* [1,2,3,7,8,10,11,12] > *Output:* [[1,2,3],[7,8],[10,11,12]] > > I have written the following code and it does the trick. > > -- Take a list and divide it at first point of non-consecutive number > encounter > divide :: [Int] -> [Int] -> ([Int], [Int]) > divide first [] = (first, []) > divide first second = if (last first) /= firstSecond - 1 then (first, > second) > else divide (first ++ [firstSecond]) (tail second) > where firstSecond = head second > > > -- Helper for breaking a list of numbers into consecutive sublists > breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] > breakIntoConsecsHelper [] [[]] = [[]] > breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] > else ans ++ [one] ++ > breakIntoConsecsHelper two [] > where > firstElem = head lst > remaining = tail lst > (one, two) = divide [firstElem] > remaining > > > -- Break the list into sublists of consective numbers > breakIntoConsecs :: [Int] -> [[Int]] > breakIntoConsecs lst = breakIntoConsecsHelper lst [[]] > > -- Take the tail of the result given by the function above to get the > required list of lists. > > However, I was wondering if there was a better way of doing this. Any help > would be highly appreciated. > > Thank you. > Best Regards, > Saqib Shamsi >

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Better Code
by Saqib Shamsi 13 Jan '17

13 Jan '17

Hi, The problem that I wish to solve is to divide a (sored) list of integers into sublists such that each sublist contains numbers in consecutive sequence. For example, *Input:* [1,2,3,7,8,10,11,12] *Output:* [[1,2,3],[7,8],[10,11,12]] I have written the following code and it does the trick. -- Take a list and divide it at first point of non-consecutive number encounter divide :: [Int] -> [Int] -> ([Int], [Int]) divide first [] = (first, []) divide first second = if (last first) /= firstSecond - 1 then (first, second) else divide (first ++ [firstSecond]) (tail second) where firstSecond = head second -- Helper for breaking a list of numbers into consecutive sublists breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] breakIntoConsecsHelper [] [[]] = [[]] breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] else ans ++ [one] ++ breakIntoConsecsHelper two [] where firstElem = head lst remaining = tail lst (one, two) = divide [firstElem] remaining -- Break the list into sublists of consective numbers breakIntoConsecs :: [Int] -> [[Int]] breakIntoConsecs lst = breakIntoConsecsHelper lst [[]] -- Take the tail of the result given by the function above to get the required list of lists. However, I was wondering if there was a better way of doing this. Any help would be highly appreciated. Thank you. Best Regards, Saqib Shamsi

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Re: [Haskell-beginners] Better Code
by Carlo Matteo Scalzo 13 Jan '17

13 Jan '17

Hi Saqib, perhaps something like this? divide :: [Int] -> [[Int]] divide [] = [] divide (x:xs) = divideAux xs [] [x] x divideAux :: [Int] -> [[Int]] -> [Int] -> Int -> [[Int]] divideAux [] result current max = result ++ [current] divideAux (x:xs) result current max = if x - max > 1 then divideAux xs (result ++ [current]) [x] x else divideAux xs result (current ++ [x]) x This can of course be generalised to other types (as per Joel's requirement) just by replacing the check on line 7, i.e., 'x - max > 1'. Happy to explain/talk about the code if you'd like me to :-) Cheers, Carlo On Fri, Jan 13, 2017 at 7:14 PM, Joel Neely <joel.neely(a)gmail.com> wrote: > Had a chance to chat with ghci, so earlier conjecture not confirmed: > > Prelude Data.List> groupBy (\x y -> x == y-1) [1,2,3,7,8,10,11,12] > > [[1,2],[3],[7,8],[10,11],[12]] > > So close but no cigar. > > On Fri, Jan 13, 2017 at 10:05 AM, Saqib Shamsi <shamsi.saqib(a)gmail.com> > wrote: > >> Hi, >> >> The problem that I wish to solve is to divide a (sored) list of integers >> into sublists such that each sublist contains numbers in consecutive >> sequence. >> >> For example, >> *Input:* [1,2,3,7,8,10,11,12] >> *Output:* [[1,2,3],[7,8],[10,11,12]] >> >> I have written the following code and it does the trick. >> >> -- Take a list and divide it at first point of non-consecutive number >> encounter >> divide :: [Int] -> [Int] -> ([Int], [Int]) >> divide first [] = (first, []) >> divide first second = if (last first) /= firstSecond - 1 then (first, >> second) >> else divide (first ++ [firstSecond]) (tail second) >> where firstSecond = head second >> >> >> -- Helper for breaking a list of numbers into consecutive sublists >> breakIntoConsecsHelper :: [Int] -> [[Int]] -> [[Int]] >> breakIntoConsecsHelper [] [[]] = [[]] >> breakIntoConsecsHelper lst ans = if two == [] then ans ++ [one] >> else ans ++ [one] ++ >> breakIntoConsecsHelper two [] >> where >> firstElem = head lst >> remaining = tail lst >> (one, two) = divide [firstElem] >> remaining >> >> >> -- Break the list into sublists of consective numbers >> breakIntoConsecs :: [Int] -> [[Int]] >> breakIntoConsecs lst = breakIntoConsecsHelper lst [[]] >> >> -- Take the tail of the result given by the function above to get the >> required list of lists. >> >> However, I was wondering if there was a better way of doing this. Any >> help would be highly appreciated. >> >> Thank you. >> Best Regards, >> Saqib Shamsi >> >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners >> >> > > > -- > Beauty of style and harmony and grace and good rhythm depend on > simplicity. - Plato > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > > -- Carlo Matteo Scalzo Mobile: +44 (0) 7934 583582 E-mail: cmscalzo(a)gmail.com

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Runtime error while feeding a binary to stdin
by Manuel Vázquez Acosta 11 Jan '17

11 Jan '17

Hi all, I'm quite new to Haskell. While following the "Real World Haskell" and doing some experimentation I came up with a anoying situation: Trying to read data from stdin it seems that binary data is not allowed. A simple "copy" program: -- file: copy.hs import System.IO main = do input <- hGetContents stdin hPutStr input Fails when I run it like: $ ghc copy.hs $ ./copy < input > output copy: <stdin>: hGetContents: invalid argument (invalid byte sequence) input contains binary data. In fact of all the following programs only the first works with binary data: copy:: IO () copy = do bracket (openBinaryFile "input" ReadMode) hClose $ \hi -> do bracket (openBinaryFile "ouput" WriteMode) hClose $ \ho -> do input <- hGetContents hi hPutStr ho input copy2:: IO () copy2 = do -- Doesn't work with binary files source <- readFile "input" writeFile "output" source copy3:: IO () copy3 = do -- Doesn't work with binary files either interact (map $ \x -> x) copy4:: IO () copy4 = do input <- hGetContents stdin hPutStr stdout input But I lost any chance of piping and/or using '<', '>' in the shell. Best regards, Manuel.

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dynamic code compilation and loading
by Dennis Raddle 09 Jan '17

09 Jan '17

Can I get a pointer to how to do dynamic code compilation and loading into a running program? D

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Autogenerating tuple constructor of length n
by Manikanda Krishnan 09 Jan '17

09 Jan '17

Hi, I am exploring haskell. Is there a way to generate a tuple constructor of n elements at runtime? Basically I want to convert a list of lists into a list of n tuples. Thanks in advance Mani

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