- Beginners - Haskell.org

Syntax error in input
by bahadýr altan 02 Apr '12

02 Apr '12

Hi, I wrote that function: wordCount x = length(words x) but I get this : ERROR "a.hs":73 - Syntax error in input (unexpected symbol "wordCount") Can you tell me how can I get rid of this error please? Thanks..

1 1

Type error in function binding
by bahadýr altan 02 Apr '12

02 Apr '12

Hi. I'm trying to write a code which counts the number of 'a's in a string. it looks like this : mycounter st = if st == [] then 0 else if head st == 'a' then 1 + mycounter tail st else mycounter tail st but it gives this error : ERROR "deneme.hs":19 - Type error in function binding *** Term :mycounter *** Type : [Char] -> b *** Does not match : ([a] -> [a]) -> [Char] -> b *** Because : unification would give infinite type I have no idea about what I'm doing wrong. Can you help me with this please?

4 3

Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35
by franco00＠gmx.com 30 Mar '12

30 Mar '12

Indeed the second snipper contains quite an obvious mistake. Thanks for noticing! It doesn't seem to me it utilises a lambda expression though? You mean the '.' operator for chaining function? If that's it, it could be rewritten unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique (filter (/= x) xs) | otherwise = x : unique xs ----- Original Message ----- From: Ramesh Kumar Sent: 03/28/12 10:14 AM To: franco00(a)gmx.com, beginners(a)haskell.org Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35 Thanks Franco, Your (first) solution is the only one which has worked so far although it utilizes a lambda expression. The problem is indeed tricky. ----------------------------------------------------------------- From: "franco00(a)gmx.com" <franco00(a)gmx.com> To: beginners(a)haskell.org Sent: Wednesday, March 28, 2012 3:39 PM Subject: Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35 gah sorry I obviously meant to reply to the "Unique integers in a list" message ----- Original Message ----- From: franco00(a)gmx.com Sent: 03/28/12 09:36 AM To: beginners(a)haskell.org Subject: Re: Beginners Digest, Vol 45, Issue 35 unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = (unique . filter (/= x)) xs | otherwise = x : unique xs -- This is a simpler to read version (albeit inefficient?) unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs _______________________________________________ Beginners mailing list Beginners(a)haskell.org http://www.haskell.org/mailman/listinfo/beginners

4 6

Non ghc specific FRP (using uhc js backend)
by Nathan Hüsken 29 Mar '12

29 Mar '12

Hi, I am currently informing myself about FRP. Looking at the Elerea paper (http://sgate.emt.bme.hu/documents/patai/publications/PataiWFLP2010.pdf) I read that it is ghc specific because of mutable references. I am trying to use the js backend of uhc and that does not support this. I have tried to use other FRP libraries with uhc (by compiling a file which does nothing but import the library) and all failed because of something missing in Control.Concurrent. So I am wondering: Is it a property FRP that it needs something like mutable references for an efficient implementation. Is there a FRP library that can be used (or can be changed so that it can be used) with uhc and its js backend? Thanks! Nathan

3 3

given these types (HFuse), can I share state?
by Jon Dowland 28 Mar '12

28 Mar '12

Hi, Firstly, this is my first post to this list, I consider myself a Haskell beginner but this question might be a bit esoteric so please let me know if I'm better directing this elsewhere. I'm writing a small fuse-based filesystem[1] using hfuse[2]. I get to write functions to a defined type. One of these is 'open', for when the user opens a file; another is 'read', for when a user invokes the read syscall. Here are the type signatures: fuseOpen :: FilePath -> OpenMode -> OpenFileFlags -> IO (Either Errno fh) fuseRead :: FilePath -> fh -> ByteCount -> FileOffset -> IO (Either Errno ByteString) In the case of my filesystem, for a typical work pattern of open, read, read, read…; there's an amount of setup that is required up front (possibly combining two or more files via patches; uncompressing files, etc.) which can be quite expensive. With my current implementation, I am having to do this work in the read function[3], because I can't figure out a way of doing it in the open function and sharing the state with subsequent reads. I think this is likely to be a performance issue and I'd prefer to have it done once, in the open function. I know that there are various 'state' monads, although I've never used them. In my case the signature for my functions is fixed, but the 'return' type is the IO monad. Is there any possibility for me to hide/share state 'through' the IO monad? Thanks for your help, [1] http://jmtd.net/software/rdifffs/ [2] http://hackage.haskell.org/packages/archive/HFuse/0.2.4/doc/html/System-Fus… [3] the work takes place within 'rdiffIncrementReadFile', here: https://github.com/jmtd/rdifffs/blob/master/rdifffs.lhs#L558 -- Jon Dowland

3 2

Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35
by franco00＠gmx.com 28 Mar '12

28 Mar '12

gah sorry I obviously meant to reply to the "Unique integers in a list" message ----- Original Message ----- From: franco00(a)gmx.com Sent: 03/28/12 09:36 AM To: beginners(a)haskell.org Subject: Re: Beginners Digest, Vol 45, Issue 35 unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = (unique . filter (/= x)) xs | otherwise = x : unique xs -- This is a simpler to read version (albeit inefficient?) unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs

2 1

Re: [Haskell-beginners] Beginners Digest, Vol 45, Issue 35
by franco00＠gmx.com 28 Mar '12

28 Mar '12

unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = (unique . filter (/= x)) xs | otherwise = x : unique xs -- This is a simpler to read version (albeit inefficient?) unique :: [Integer] -> [Integer] unique [] = [] unique (x:xs) | elem x xs = unique xs | otherwise = x : unique xs

1 0

Unique integers in a list
by Ramesh Kumar 28 Mar '12

28 Mar '12

Hi, I've just started learning Haskell a couple of weeks ago using Simon Thompson's "Haskell: Craft of Functional Programming". There is an exercise in chapter 7 of the book which goes something like this: Define a function of the type: unique :: [Integer] -> [Integer] which if given a list of integers, should return a list of those integers which occur only once in the input list. Example: unique [5,2,4,2,3,1,5,2] should result in [4,3,1] *** The questions assumes we know only of list comprehensions and recursion. I am guessing the solution must include something like this: unique :: [Integer] -> [Integer] unique xs = [ x | x <- xs, isSingle x ] My problem is in defining the function 'isSingle'. I would greatly appreciate any pointers on this. Many thanks. Ramesh

3 4

Unique integers in a list
by AbdulSattar Mohammed 28 Mar '12

28 Mar '12

On Wed, Mar 28, 2012 at 11:10 AM, Prasanna K Rao <prasannakrao(a)yahoo.com>wrote: > Hi, > > One way is to define a 'isin' function like this:: > > isin x (a:[]) = if (x == a) then True else False > isin x (a:as) = if (x == a) then True else isin x as > > Unnecessary. We have elem for that. > and use it like this:: > > unique (x:xs) = if not(isin x xs) then [x] ++ unique(xs) else unique(xs) > It removes duplicates. Does not remove the elements that have duplicates (which was asked by OP). It also needs the empty list check to terminate. It'll fail with Non-exhaustive patterns. > > or like this:: > > unique(x:xs) = [x | x <- (x:xs), not(isin x xs)] ++ unique xs > The not(isin x xs) will definitely fail for all the elements except the first one because those elements are being taken from xs. > > The later being the preferred one. HTH.. > > If you see correctly, the former is the preferred one (giving different solution). To OP, When we pass x to isSingle x xs, we know that there is at least one x in xs. If we remove that x and check for the existence of x in the remainder of the list, we know if there is more than one x. isSingle x xs = x `notElem` (delete x xs) delete is in Data.List. > Regards, > > > ------------------------------ > *From:* Ramesh Kumar <rameshkumar.techdynamics(a)ymail.com> > *To:* "Beginners(a)haskell.org" <Beginners(a)haskell.org> > *Sent:* Wednesday, March 28, 2012 3:03 AM > *Subject:* [Haskell-beginners] Unique integers in a list > > Hi, > > I've just started learning Haskell a couple of weeks ago using Simon > Thompson's "Haskell: Craft of Functional Programming". > There is an exercise in chapter 7 of the book which goes something like > this: > > Define a function of the type: unique :: [Integer] -> [Integer] > which if given a list of integers, should return a list of those integers > which occur only once in the input list. > Example: > unique [5,2,4,2,3,1,5,2] should result in [4,3,1] > > > *** The questions assumes we know only of list comprehensions and > recursion. > > I am guessing the solution must include something like this: > > unique :: [Integer] -> [Integer] > unique xs = [ x | x <- xs, isSingle x ] > > My problem is in defining the function 'isSingle'. > > I would greatly appreciate any pointers on this. > > Many thanks. > Ramesh > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > -- Warm Regards, AbdulSattar Mohammed -- Warm Regards, AbdulSattar Mohammed

2 1

Haskell wants the type, but I only know the class.
by Amy de Buitléir 26 Mar '12

26 Mar '12

I'm trying to read something from a file, do something with it, and then write it out again. I don't know, or care about, the type of the object I'm reading, but I do know its class. Here's my code: ---------- import Data.Binary ( Binary, encode, decode ) import Data.ByteString.Lazy as B ( readFile, writeFile ) import Codec.Compression.GZip ( compress, decompress ) class ( Binary a, Show a, Eq a ) => Thing a where doSomething :: a -> IO a readThing :: ( Thing a ) => FilePath -> IO a readThing f = return . decode . decompress =<< B.readFile f writeThing :: ( Thing a ) => FilePath -> a -> IO () writeThing f = B.writeFile f . compress . encode main = do a <- readThing "file1.txt" a' <- doSomething a writeThing "file2.txt" a' ---------- And here's the compiler error I get. Amy.hs:18:3: Ambiguous type variable `a0' in the constraint: (Thing a0) arising from a use of `writeThing' Probable fix: add a type signature that fixes these type variable(s) In the expression: writeThing "file2.txt" a' In the expression: do { a <- readThing "file1.txt"; a' <- doSomething a; writeThing "file2.txt" a' } In an equation for `main': main = do { a <- readThing "file1.txt"; a' <- doSomething a; writeThing "file2.txt" a' } Is there a way to fix my code? I tried adding a type signature, e.g., a <- readThing "file1.txt" :: (Thing t) => IO t a' <- doSomething a :: (Thing t) => IO t but then it says... Couldn't match type `t0' with `t' because type variable `t' would escape its scope This (rigid, skolem) type variable is bound by an expression type signature: Thing t => IO t The following variables have types that mention t0 a :: t0 (bound at Amy.hs:16:3) In a stmt of a 'do' expression: a' <- doSomething a :: Thing t => IO t In the expression: do { a <- readThing "file1.txt" :: Thing t => IO t; a' <- doSomething a :: Thing t => IO t; writeThing "file2.txt" a' } In an equation for `main': main = do { a <- readThing "file1.txt" :: Thing t => IO t; a' <- doSomething a :: Thing t => IO t; writeThing "file2.txt" a' }

5 16