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Re: [Haskell-beginners] Simplified Luhn Algorithm
by mrx 31 Dec '17

31 Dec '17

I guess your issue is how to represent the card numbers of arbitrary length, or? Wouldn't a list work? Patrik Den 31 dec 2017 05:03 skrev "trent shipley" <trent.shipley(a)gmail.com>: I have the following, and it works, but I am trying teach myself Haskell, and I have the suspicion that my solutions is both inefficient and graceless. Any feedback would be appreciated. Trent. ------------------------------------ {- 8.The Luhn algorithm is used to check bank card numbers for simple errors such as mistyping a digit, and proceeds as follows: * consider each digit as a separate number; * moving left, double every other number from the second last; * subtract 9 from each number that is now greater than 9; * add all the resulting numbers together; * if the total is divisible by 10, the card number is valid. Define a function luhnDouble :: Int -> Int that doubles a digit and subtracts 9 if the result is greater than 9. For example: > luhnDouble 3 6 > luhnDouble 6 3 Using luhnDouble and the integer remainder function mod, define a function luhn :: Int -> Int -> Int -> Int -> Bool that decides if a four-digit bank card number is valid. For example: > luhn 1 7 8 4 True > luhn 4 7 8 3 False In the exercises for chapter 7 we will consider a more general version of this function that accepts card numbers of any length. Hutton, Graham. Programming in Haskell (pp. 45-46). Cambridge University Press. Kindle Edition. -} luhnDouble :: Int -> Int luhnDouble x = if (2 * x) > 9 then (2 * x) - 9 else 2 * x luhn :: Int -> Int -> Int -> Int -> Bool luhn x1 x2 x3 x4 = if 0 == sum[luhnDouble x1, x2, luhnDouble x3, x4] `mod` 10 then True else False _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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type family exercise
by info＠maximka.de 29 Dec '17

29 Dec '17

There is a little exercise at the end of the post "Type Tac Toe: Advanced Type Safety" http://chrispenner.ca/posts/type-tac-toe > As an exercise try combining playX and playO into a more general play! Here's a hint, you'll want to make another wrapper type like we did with Coord! Link to the code https://github.com/ChrisPenner/Type-Tac-Toe/blob/master/src/TypeTacToe.hs Unfortunately I couldn't solve it by myself. The only generalised play I could implement is here: play :: (Played x y b ~ 'False) => (Coord x, Coord y) -> Board b PieceT -> PieceT -> Board b PieceT play (coordVal -> x, coordVal -> y) (Board b) p = Board $ overTrip y (overTrip x (const p)) b I would be very grateful for any help to solve the exercise by using type family extension. Regards, Alexei

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function multiple args
by mike h 19 Dec '17

19 Dec '17

Hi, I have a number of functions like these: v :: Int -> [a] -> [a] w :: Int -> Int -> [a] -> [a] x :: a -> a -> [a] -> [a] y :: a -> b ->… <other args>... -> [a] -> [a] z… etc. where there are any number of args of different types but the last two are common to all the functions i.e. they all have [a] -> [a] What I’m trying to do is build a collection (ordered) of such functions and then apply each in turn to a [a] to get the final [a]. What would be the best approach to this? Thanks Mike

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Re: [Haskell-beginners] Int and Integer
by Peter Normington 17 Dec '17

17 Dec '17

Thanks, Theodore. As simple as that! Best regards, Peter On 17 December 2017 at 15:40, Theodore Lief Gannon <tanuki(a)gmail.com> wrote: > Because: > (!!) :: [a] -> Int -> a > > On Dec 17, 2017 5:50 AM, "Peter Normington" <peter(a)ireby.org.uk> wrote: > >> Why do I get an error in the following example: >> >> memoized_fib :: Integer -> Integer >> memoized_fib = (map fib [0 ..] !!) >> where fib 1 = 1 >> fib 2 = 1 >> fib n = memoized_fib (n-2) + memoized_fib (n-1) >> >> but not in: >> >> memoized_fib :: Int -> Integer >> memoized_fib = (map fib [0 ..] !!) >> where fib 1 = 1 >> fib 2 = 1 >> fib n = memoized_fib (n-2) + memoized_fib (n-1) >> ? >> >> The error is reported by GHCi as: >> >> error: >> • Couldn't match type ‘Integer’ with ‘Int’ >> Expected type: Integer -> Integer >> Actual type: Int -> Integer >> • In the expression: (map fib [0 .. ] !!) >> In an equation for ‘memoized_fib’: >> memoized_fib >> = (map fib [0 .. ] !!) >> where >> fib 1 = 1 >> fib 2 = 1 >> fib n = memoized_fib (n - 2) + memoized_fib (n - 1) >> | >> 2 | memoized_fib = (map fib [0 ..] !!) >> | ^^^^^^^^^^^^^^^^^ >> Failed, 0 modules loaded. >> >> >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners >> >>

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Int and Integer
by Peter Normington 17 Dec '17

17 Dec '17

Why do I get an error in the following example: memoized_fib :: Integer -> Integer memoized_fib = (map fib [0 ..] !!) where fib 1 = 1 fib 2 = 1 fib n = memoized_fib (n-2) + memoized_fib (n-1) but not in: memoized_fib :: Int -> Integer memoized_fib = (map fib [0 ..] !!) where fib 1 = 1 fib 2 = 1 fib n = memoized_fib (n-2) + memoized_fib (n-1) ? The error is reported by GHCi as: error: • Couldn't match type ‘Integer’ with ‘Int’ Expected type: Integer -> Integer Actual type: Int -> Integer • In the expression: (map fib [0 .. ] !!) In an equation for ‘memoized_fib’: memoized_fib = (map fib [0 .. ] !!) where fib 1 = 1 fib 2 = 1 fib n = memoized_fib (n - 2) + memoized_fib (n - 1) | 2 | memoized_fib = (map fib [0 ..] !!) | ^^^^^^^^^^^^^^^^^ Failed, 0 modules loaded.

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Coredump functionality for Haskell
by Baa 15 Dec '17

15 Dec '17

Hello, All! In C if I need I can abort program and generate coredump - to research state of stack/vars. Is it possible in Haskell - to abort and to generate some dump which can be easy explored/researched in debugger with Haskell symbolic info (not standard Linux coredump)? And to generate it without any coding/hacking? === Best regards, Paul

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Why exception is not catching?!
by Baa 15 Dec '17

15 Dec '17

Hello, dear List! Super simple function - "safe enum conversion". It's based on `Either` because only it instantiates `MonadCatch` (Maybe does not, but with MaybeT I've got error that internal monad Indetity is not MonadCatch, so I avoid to use Maybe). That it is: data A = A1|A2 deriving (Enum, Show) data B = B1|B2|B3 deriving (Enum, Show) conv :: B -> Either SomeException A conv b = safeConv where catchError (e::SomeException) = Left e safeConv = (Right $ toEnum $ fromEnum b) `catch` catchError -- toRes (Left _) = Nothing -- toRes (Right r) = Just r But exception is not catching and I'm getting it in GHCI repl :) Exceptions are different?! What's wrong with this code? === Best regards, Paul

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How to install new GHC ?
by Baa 13 Dec '17

13 Dec '17

Hello, All, I want to switch to the new GHC (8.2.2) because I hit some bug in GHC and suppose that it may be fixed in the new version. So, to try it I changed "resolver: XXX" in my .yaml-files to "lts-9.18". Also I done something else (I don't remember what exactly) and call `stack setup ... --reinstall ...`. New GHC have been installed. And I switched to new LTS - I see it 100%. But when I run `stack build`, IMHO compiller is old, because if I call `stack ghci` I see that old GHC (8.0.2) is calling. I'm not familiar with stack/Haskell tools in-depth and may miss something! For example, I know that LTS-9.18 is last stable, GHC 8.2.2 is last too, but in Stackage seems that 9.18 corresponds to 8.2.2. If I run `stack setup` in the directory with my project again I get: --cut-- stack will use a sandboxed GHC it installed For more information on paths, see 'stack path' and 'stack exec env' To use this GHC and packages outside of a project, consider using: stack ghc, stack ghci, stack runghc, or stack exec --cut-- Would somebody explain me what is wrong here and how I can test a project with new 8.2.2 GHC? === Best regards, Paul

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Re: [Haskell-beginners] IO question
by Vale Cofer-Shabica 12 Dec '17

12 Dec '17

Just to add to David's answer, when I find myself in situations like: foo >>= return . bar hlint helpfully suggests: fmap bar foo So you could also have: fmap (splitOn ",") $ readFile "data.txt" -vale -- vale cofer-shabica 401.267.8253 On Mon, Dec 11, 2017 at 4:24 PM, mike h <mike_k_houghton(a)yahoo.co.uk> wrote: > Thank you David. Before posting I tried readFile "data.txt" >>= … but got > errors as I didn’t use return.!! > > Mike > > > > On 11 Dec 2017, at 21:00, David McBride <toad3k(a)gmail.com> wrote: > > splitOn "," <$> readFile "data.txt" > > or perhaps > > readFile "data.txt" >>= return . splitOn "," > > On Mon, Dec 11, 2017 at 3:56 PM, mike h <mike_k_houghton(a)yahoo.co.uk> > wrote: > >> I have >> >> input <- readFile “data.txt” >> let input’ = splitOn “,” input >> …. >> >> How do I make that into just one line? >> >> Thanks >> >> Mike >> _______________________________________________ >> 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 > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > >

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IO question
by mike h 11 Dec '17

11 Dec '17

I have input <- readFile “data.txt” let input’ = splitOn “,” input …. How do I make that into just one line? Thanks Mike

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