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Re: Problems with IO
by ?????? ????? ????? ?????? 24 Sep '02

24 Sep '02

Now I understand that it's just a lack of knowledge... Sorry for interrupring... By the way, can you recommend a good manual which I can find on the web? -- Zhbanov Pavel

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Problems with IO
by Жбанов Павел Геннад ьевич 24 Sep '02

24 Sep '02

I'm having a problem with writing a function dealing with I/O. Maybe it's just a lack of experience or simple Haskell knowledge because I'm just a beginer. The problem: I want to write a function that converts an IO String into String lala :: IO String -> String Is it possible? If yes, how? -- Жбанов Павел

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question regarding ordering of function definitions
by David Roundy 23 Sep '02

23 Sep '02

Hello all! I've just started learning (and using haskell), and am also using the literate comment mode (latex style), and just ran into a problem that seems like it ought to have a simple solution, which is that it seems that haskell requires you to define all the variations of a function sequentially, which is raining on my literate programming parade. Here is what I'd like to do: \begin{code} data D = A String | B String instance Show D where show = showD instance Read D where readsPrec _ = readsD readD s = (readsA s) ++ (readsB s) \end{code} A is formatted like ``A string''. \begin{code} showD (A s) = "A " ++ s readsA s = [(A thestr, r) | ("A", x) <- mylex s, (thestr, r) <- mylex x] \end{code} B is formatted like ``B string''. \begin{code} showD (B s) = "B " ++ s readsB s = [(B thestr, r) | ("B", x) <- mylex s, (thestr, r) <- mylex x] \end{code} The problem I have is that ghc complains because I've split up the definition of showD with readsA defined in between. This surprised me, since it seemed that the code should still be unambiguous. Is there some nice way around this, or do I have to define a separate function for each constructor, and then have a list in one place (as I do with the readsD, in that case because pattern matching won't give me what I want), like: showD (A s) = showA (A s) showD (B s) = showB (B s) Of course, in my actual example, there are more than two constructors, and the parsing is considerably more complex, which is why I'd like to put the analagous showing and parsing right next to eachother along with a description of what the formatting is intended to be. -- David Roundy http://civet.berkeley.edu/droundy/

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"zipping" two streams in the IO monad
by Nick Name 20 Sep '02

20 Sep '02

How could I "zip" two possibily infinite streams obtained by IO operations, for example to return the infinite list of pairs of values read both from channel a and from channel b with getChanContents? In general, how does one implement a function like "getChanContents" wich returns an infinite list? Tell me just some pointer to information of course, I don't want an entire functional programming lesson :) Vincenzo

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Depth-First search problems
by Dennis Schieferdecker 20 Sep '02

20 Sep '02

Hi! I hope you can help me with this little problem I have: I'm trying to implement a depth-first search on a binary tree from a school book but somehow I always get Haskell error messages (more after the source code). The source is as follows: -- Binary Tree structure data BinTree t = EmptyTree | Bin (BinTree t) t (BinTree t) left (Bin b1 _ _) = b1 right (Bin _ _ b2) = b2 value (Bin _ v _) = v empty EmptyTree = True empty (Bin _ _ _) = False -- Give the Binary Tree == instance Eq a => Eq (BinTree a) where (Bin b v c) == (Bin d w e) = (v == w) && (b == d) && (c == e) EmptyTree == EmptyTree = True EmptyTree == Bin _ _ _ = False -- Stack structure data Keller t = CreateStack | Push (Keller t) t top (Push s x) = x pop (Push s x) = s -- Depth-First search tiefensuche b = fst (until p f ([], Push CreateStack b)) where p(_, k) = empty k f(erg, k) | (top k) == EmptyTree = (erg, pop k) | otherwise = ([v] ++ erg, Push( Push( pop k) b2) b1) where (Bin b1 v b2) = top k The error message says that p has a wrong types Haskell thinks it is p :: (b, BinTree c) -> Bool, instead of p :: ([a], Keller (BinTree a)) -> Bool I really don't know how I can tell to see the right type. Does anyone have an idea what I have to change ? -- ciao dennis "I hear and I forget, I see and I remember, I do and I understand"

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layout problem
by Γιώργος Κοσμίδης 18 Sep '02

18 Sep '02

Hello everyone. I have a problem with a assessment. The problem is with the layout. I cannot understand what is wrong. Here is the code module Main where import IO import System import List import Maybe import Char import Numeric type Word=String type WordCount=Int type WordTup=(word,wordCount) type WordList=[WordTup] main=do args<-getArgs textLines<-getText switchArgs args textLines where switchArgs args textLines |(isMemberOf "-f" args) && (isMemberOf "-i" args) =printFreq(sortWordCount(parseLcLine textLines)) |(isMemberOf "-f" args) =printFreq(sortWordCount (parseLine textLines)) |(isMemberOf "-i" args) =printFreq(sortWordCount (parseLcLine textLines)) |otherwise =printFreq(sortName (parseLine textLines)) getText::IO String getText=do c<-getChar e<-isEOF if (e) then return "" else (do nc<-getText return (:nc)) parseLine::String->Wordlist parseLine line |(line/="")&&(isWantedChar (headline)) =fillWordList w(parseLine newLine) |(line/="")&&(not(isWantedChar (headline))) =parseLine(tail line) isMemberOf::String->[String]->Bool isMemberOf arg args =[]/=[a|a<-args,a==arg] printFreq::WordList->IO() printFreq wl=do sequence (map putWordStat wl) where putWordStat(Word,WordCount)->IO() putWordStat(w,wc) do putStr w putSpc (20 (length w) (length(show wc))) putStr (show wc) putSpc 8 putStr (showFFloat (Just 2) (fromIntegral (wc) * 100.0/fromIntegral (wLength))"") putChar '%' putChar '\n' return() wlLength-countWords wl; putSpc::Int->IO() putSpc n |n>1 do putChar '' putSpc (n-1) return() |otherwise=putChar '' countWords:Wordlist->Int countWords [ ] countWords ((w,wc):wl)=wc+countWords wl sortName::WordList->WordList sortName [ ] = [ ] sortName (wn:ws)=sortName [(w,wc) (w,wc) <-ws,not (isGreater (w,wc)wc)] ++[wn]++sortName[(w,wc)|(w,wc)<-ws, isGreater (w,wc) wn] where isGreater (wl,wcl) (w2,wc2) |ord(head(wl))-ord(head(x2))>0=True |ord(head(wl))-ord(head(x2))<0=False |length wl>l && length w2>l =isGreater(tail(wl),wcl) (tail(w2,wc2) |otherwise = [] where w=parseWord line newline=drop (length w) line parseLcLine::String->WordList parseLcLine line |(line/="") && (isWantedChar (head line))=fillWordList w (parseLcLine newline) |(line/="") && (not (isWantedChar (head line)))=parseLcLine (tail line) |otherwise = [] where w=map toLower (parseWord line) newline=drop (lenfth w) line parseWord::Word->Word parseWord w |(w/="") && (isWantedChar (head w))=(head w): parseWord (tail w)) |otherwise = "" isWantedChar::Char->Bool isWantedChar c | (c=='a'|| c=='b'|| c=='c'|| c=='d'|| c=='e'|| c=='f'|| c=='g'|| c=='h'|| c=='i'|| c=='j'|| c=='k'|| c=='l'|| c=='m'|| c=='n'|| c=='o'|| c=='p'|| c=='q'|| c=='r'|| c=='s'|| c=='t'|| c=='u'|| c=='v'|| c=='w'|| c=='x'|| c=='y'|| c=='z'||c=='A'|| c=='B'|| c=='C'|| c=='D'|| c=='E'|| c=='F'|| c=='G'|| c=='H'|| c=='I'|| c=='J'|| c=='K'|| c=='L'|| c=='M'|| c=='N'|| c=='O'|| c=='P'|| c=='Q'|| c=='R'|| c=='S'|| c=='T'|| c=='U'|| c=='V'|| c=='W'|| c=='X'|| c=='Y'|| c=='Z'|| = True) |otherwise = False fillWordList::Word->WordList->WordList fillWordList w wordlistl |cWord/= [] =cWord++[wordl,wordCountl)|(wordl,wordCountl<-wordListl,wordl/=w] |otherwise = (w,l):wordListl where cWord=[(wordl,(wordCountl-l)) | (wordl,wordCountl)<-wordListl,wordl==w] |length w1>1 && length w2==1=True |otherwise = False sortWordCount::WordList->WordList sortWordCount [] = [] sortWordCount ((wn,wcn):ws=sortWordCount [(w,wc|cw,wc)<-ws,ws<=wvn++ [(wn,wcn)]++sortWordCount [(cw,wc)|(w,wc)<-ws,wcs,wcn]

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(no subject)
by Richard Barth 14 Sep '02

14 Sep '02

I'm just learning Haskell with Hugs98. I can't understand why the line defining "foo" is acceptable while the line defining "bar" (only difference is the unary minus) generates the error message at the end. If someone could explain this, or, better yet, tell me how I can figure this out for myself. Thanks. data Entry = Ent {date :: String, amount :: Double} foo = [Ent "baz" 1.0] bar = [Ent "baz" -1.0] Type checking...... ERROR "...test.hs":7 - Unresolved top-level overloading *** Binding : bar *** Outstanding context : Fractional (Double -> Entry)

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Re: (no subject)
by Ashley Yakeley 14 Sep '02

14 Sep '02

At 2002-09-13 12:12, Richard Barth wrote: >I'm just learning Haskell with Hugs98. I can't understand why the line >defining "foo" is acceptable while the line defining "bar" (only >difference is the unary minus) generates the error message at the end. >If someone could explain this, or, better yet, tell me how I can figure >this out for myself. Thanks. > >data Entry = Ent {date :: String, amount :: Double} >foo = [Ent "baz" 1.0] >bar = [Ent "baz" -1.0] Haskell thinks it's a binary minus. Try this: bar = [Ent "baz" (-1.0)] -- Ashley Yakeley, Seattle WA

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maxList
by Nuno Silva 13 Sep '02

13 Sep '02

hello, I'm new to haskell, I'm studying for an exam, and I'm having trouble with this exercise. I have a list of Integers, and I want to return the highest value of all. i have this: comp :: Int -> Int -> Int comp x y | x < y = y | otherwise = x maxList :: [Int] -> [Int] maxList (x1:x2:xs) = comp x1 x2 : maxList xs maxList [] = [] this code returns a list, but I only want it to return a single Int... but can't seem to do it, could you please help me? --------------------------------- Do you Yahoo!? Yahoo! News - Today's headlines

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by Your Friend 13 Sep '02

13 Sep '02

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