January 2009 - Beginners

Understanding cached fibonnacci function
by Patrick LeBoutillier 31 Jan '09

31 Jan '09

Hi all, I recently stumbled on this example in some wiki: mfib :: Int -> Integer mfib = (map fib [0 ..] !!) where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) I don't understand how the results get cached. When mfib is recursively called, doesn't a new (map fib [0 ..] !!) start over again? Or perhaps I'm thinking too imperatively here... Also, if I change the definition to this (adding "a" on both sides): mfib :: Int -> Integer mfib a = (map fib [0 ..] !!) a where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) the funtion becomes slow again. Why is that? Thanks a lot, Patrick LeBoutillier -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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Understanding cached fibonnacci function
by Patrick LeBoutillier 30 Jan '09

30 Jan '09

Hi all, I recently stumbled on this example in some wiki: mfib :: Int -> Integer mfib = (map fib [0 ..] !!) where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) I don't understand how the results get cached. When mfib is recursively called, doesn't a new (map fib [0 ..] !!) start over again? Or perhaps I'm thinking too imperatively here... Also, if I change the definition to this (adding "a" on both sides): mfib :: Int -> Integer mfib a = (map fib [0 ..] !!) a where fib 0 = 0 fib 1 = 1 fib n = mfib (n-2) + mfib (n-1) the funtion becomes slow again. Why is that? Thanks a lot, Patrick LeBoutillier -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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Re : Re: [Haskell-beginners] Compile type Error
by emmanuel.delaborde 30 Jan '09

30 Jan '09

On 29 Jan 2009, at 16:10, beginners-request(a)haskell.org wrote: > Content-Type: text/plain; charset="iso-8859-1" > > I didn't compile, but, by the looks of it, I would change > > import qualified Data.ByteString as BS (readFile) > > for > > import qualified Data.ByteString.Lazy as BS (readFile) > > > as it seems sha1 needs a lazy bytestring Thanks ! It compiles now but brings me to my real problem :) If I uncomment this line "md5" -> md5sum I get the following : ------ module Main where import System.Environment (getArgs) import Data.Digest.OpenSSL.MD5 (md5sum) import Data.Digest.Pure.SHA (sha1, showDigest) import qualified Data.ByteString.Lazy as BS (readFile) -- sha1 :: ByteString -> Digest -- readFile :: FilePath -> IO ByteString -- md5sum :: ByteString -> String -- showDigest :: Digest -> String checkHash :: String -> String -> String -> IO () checkHash codec hash file = let f = case codec of "md5" -> md5sum "sha1" -> showDigest . sha1 _ -> error "Codec must be md5 or sha1 !" in BS.readFile file >>= \fileBS -> let fileHash = f fileBS in print (hash == fileHash) main = getArgs >>= \as -> case as of (codec:hash:file:_) -> checkHash codec hash file _ -> usage usage = print "checksum codec hash file" ------ which now fails to compile with this error : checksum.hs:17:22: Couldn't match expected type `Data.ByteString.Internal.ByteString' against inferred type `Data.ByteString.Lazy.Internal.ByteString' In the expression: showDigest . sha1 In a case alternative: "sha1" -> showDigest . sha1 In the expression: case codec of { "md5" -> md5sum "sha1" -> showDigest . sha1 _ -> error "Codec must be md5 or sha1 !" } How can I both use md5sum and sha1 when one use a lazy ByteString end the other one does not ? Thanks E -- Emmanuel Delaborde Web Technologist Cimex 53-55 Scrutton Street, London UK, EC2A 4PJ T: +44 (0)20 7324 7780 F: +44 (0)20 7324 7781 http://www.cimex.com ----------------------------------------------------------------------------------------------- This e-mail (and any attachments) is confidential and may contain personal views which are not the views of Cimex Media Ltd and any affiliated companies, unless specifically stated. It is intended for the use of the individual or group to whom it is addressed. If you have received it in error, please delete it from your system, do not use, copy or disclose the information in any way nor act in reliance on it and please notify postmaster(a)cimex.com A company registered in England Wales. Company Number 03765711 Registered Office : The Olde Bakehouse, 156 Watling Street East, Towcester, Northants NN12 6DB This email was scanned by Postini, the leading provider in Managed Email Security.

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Question about Real World Haskell
by Alan Cameron 29 Jan '09

29 Jan '09

In Chapter 3 of Real World Haskell it is not clear to me how to get the myInfo into the ghci system there appears to be two definitions for ch03/BookStore.hs can anyone help? Alan Cameron

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Compile type Error
by emmanuel.delaborde 29 Jan '09

29 Jan '09

Hello I have the following snippet : ---------------------------------------------------------- module Main where import System.Environment (getArgs) import Data.Digest.OpenSSL.MD5 (md5sum) import Data.Digest.Pure.SHA (sha1, showDigest) import qualified Data.ByteString as BS (readFile) -- sha1 :: ByteString -> Digest -- readFile :: FilePath -> IO ByteString -- md5sum :: ByteString -> String -- showDigest :: Digest -> String checkHash :: String -> String -> String -> IO () checkHash codec hash file = let f = case codec of --"md5" -> md5sum "sha1" -> showDigest . sha1 _ -> error "Codec must be md5 or sha1 !" in BS.readFile file >>= \fileBS -> let fileHash = f fileBS in print (hash == fileHash) main = getArgs >>= \as -> case as of (codec:hash:file:_) -> checkHash codec hash file _ -> usage usage = print "checksum codec hash file" ---------------------------------------------------------- which fails to compile, this is the error I get : checksum.hs:20:19: Couldn't match expected type `Data.ByteString.Lazy.Internal.ByteString' against inferred type `Data.ByteString.Internal.ByteString' In the first argument of `f', namely `fileBS' In the expression: f fileBS In the definition of `fileHash': fileHash = f fileBS it looks like (showDigest . sha1) expects Data.ByteString.Lazy.Internal.ByteString but gets Data.ByteString.Internal.ByteString What can I do ? Thanks E. ----------------------------------------------------------------------------------------------- This e-mail (and any attachments) is confidential and may contain personal views which are not the views of Cimex Media Ltd and any affiliated companies, unless specifically stated. It is intended for the use of the individual or group to whom it is addressed. If you have received it in error, please delete it from your system, do not use, copy or disclose the information in any way nor act in reliance on it and please notify postmaster(a)cimex.com A company registered in England Wales. Company Number 03765711 Registered Office : The Olde Bakehouse, 156 Watling Street East, Towcester, Northants NN12 6DB This email was scanned by Postini, the leading provider in Managed Email Security.

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Real World Haskell book authors accepting comments?
by Alan Cameron 29 Jan '09

29 Jan '09

The online version of the book seems to have stopped accepting comments - at least from me. Alan Cameron

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Monads...
by Cory Knapp 29 Jan '09

29 Jan '09

Hello, so, I'm having a simple problem with monads: I have no idea how to actually use them. I understand the category theory (or, at least well enough to be able to explain "what is a monad"); I understand the way to declare something as a monad instance, but I just don't get how to program with them. Can anyone provide me with, or direct me towards, some simple monads and some ways of using (for example) the monadic properties of lists? Thanks, Cory

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Help! Trapped in the IO Monad!
by Erik de Castro Lopo 29 Jan '09

29 Jan '09

Hi all, I have a list of entries for a directory (FilePaths) and I'd like to partition them into files and directories using Data.List.partition: partition :: [a] -> ([a], [a]) Now, one solution is to use unsafePerformIO: splitDirFile :: [FilePath] -> ([FilePath], [FilePath]) splitDirFile paths = do partition (\p -> unsafePerformIO (doesDirectoryExist p)) paths Two questions: a) Is it possible to do this without invoking unsafePerformIO? Ie with a function signature of say: partition :: [FilePath] -> IO ([FilePath], [FilePath]) b) Exactly how unsafe is the unsafePerformIO version? Erik -- ----------------------------------------------------------------- Erik de Castro Lopo ----------------------------------------------------------------- The main confusion about C++ is that its practitioners think it is simultaneously a high and low level language when in reality it is good at neither.

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Re: Beginners Digest, Vol 7, Issue 24
by Benedikt Ahrens 29 Jan '09

29 Jan '09

hello cory, you might want to read this article: http://stefan-klinger.de/files/monadGuide.pdf it helped me find and understand the monads in haskell. ben > Message: 2 > Date: Wed, 28 Jan 2009 20:44:30 -0600 > From: Cory Knapp <thestonetable(a)gmail.com> > Subject: [Haskell-beginners] Monads... > To: Beginners(a)haskell.org > Message-ID: <4981180E.5010204(a)gmail.com> > Content-Type: text/plain; charset=ISO-8859-1; format=flowed > > Hello, so, I'm having a simple problem with monads: I have no idea how > to actually use them. I understand the category theory (or, at least > well enough to be able to explain "what is a monad"); I understand the > way to declare something as a monad instance, but I just don't get how > to program with them. Can anyone provide me with, or direct me towards, > some simple monads and some ways of using (for example) the monadic > properties of lists? > > Thanks, > Cory >

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To use random number monad inside function
by Erick González 28 Jan '09

28 Jan '09

Hi: Haskell' s way of random number generation is strange to me, but I know how to do that. I' d like to know how can I call random numbers generated on the screen inside a classic function. I mean, I need to know the way of calling random numbers (one different every time that I run) inside a function. Please, Could anybody help me? Thanks

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