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RE: writing a function to build a list of ints from user input
by Simon Marlow 06 May '03

06 May '03

> On Tue, May 06, 2003 at 05:02:56AM -0500, Galen Menzel wrote: > > > > makeList :: IO [Int] > > makeList = do s <- getLine > > case read s of > > 0 -> return [] > > x -> do xs <- makeList > > return (x:xs) > > > > I changed the if to a case so that (read s) doesn't get > called twice. > > Does the compiler not optimize away identical calls? Or is it > not allowed to, for some reason? Yes, GHC will common up the repeated expression when -O is turned on. > On the other hand, if I write a function like: > > long_list = [1..100000] ++ [1..100000] ++ [1..100000] ++ [1..100000] > > And then run a > > putStr $ unlines $ map show long_list > > I would expect this to take very little memory, which would > mean that the > [1..100000] would need to be calculated four separate times > rather than > being calculated once and stored... Well, this is one of the disadvantages of common sub-expression elimination: you have to store the result of the sub-expression between uses. It is a space/time tradeoff, and in cases like the above, a pretty poor one. But GHC does it anyway, because in most (probably all) of the cases we've seen it is a win. Cheers, Simon

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do expression
by sashan 05 May '03

05 May '03

Hi I'm trying to write a simple program to get user input and execute a function based on the input. The program follows: import IO main = do putStrLn "1) f1" putStrLn "2) f2" choice <- getLine case choice of 1 -> do putStrLn "f1" 2 -> do putStrLn "f2" _ -> do putStrLn "" When compiled with ghc I get the following error: test.hs:8: No instance for (Num String) arising from the pattern `2' at test.hs:8 In a case alternative: 2 -> do putStrLn "f2" In the case expression: case choice of 1 -> do putStrLn "f1" 2 -> do putStrLn "f2" _ -> do putStrLn "" Please can someone explain. Thanks

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Maximum value in a list
by sashan 05 May '03

05 May '03

Hi I was reading through the tutorial Yet Another Haskell Tutorial and doing chapter 3 ex 5, write a function using foldr and max to return the maximum value in a list. I came up with the following that will work for positive numbers. maxInList :: [Int]->Int maxInList [] = 0 maxInList l = foldr max 0 l Is there an identity for the max function? Because currently if the list [-1,-3,-4] is passed to maxInList it will return 0.

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Re: Fast Mutable arrays and a general question about IO
by Jon Cast 05 May '03

05 May '03

Ron de Bruijn <rondebruijn(a)yahoo.com> wrote: > > Why? It's not always necessary to go for maximum speed; do you have > > a reason for needing it in this case? > > Almost any operation in my program works on array's and before the > program terminates, there have been an awful lot of operations on > it. If there is an other kind of array that mutates and doesn't create > a copy of it when changing a value and isn't more than 1,5 times as > slow as an array in C++ then that should also be ok. The program would > run in a scale of hours or days. And if it's twice as fast, than I > could calculate two times as much. Alright. <snip> > > I think you mean `above'. In any case, could you show us some code > > you have that's not working? That'll make it much easier for us to > > answer your questions. > > > > Well I don't have really code, because I want to begin programming it. > > I will restate my goal: I want to be able to read, update and create > arrays in Haskell of type myDataType (suppose my constructor function > is Con Int String Int) Alright. You seem to have the concepts down here, so I would recommend you try to start on your own, and mail us back when you have a problem. > on a way that's fast, preferabily the fastest method. I don't know what the fastest method is, although I doubt you'll get anything (natively) faster than IOArray, or IOUArray if you only need arrays of primitive types. <snip> > > > So to put it more concrete: In any Haskell program that does IO > > > there is always some function that has this form: > > > > > > do x<-someSource > > > putBoundVariableToSomeOtherFunctionThatDoesMonadicOperations > > > (restOfPureFunctionalProgramForExample x) > > > > No. > > Why not? Sorry; I put that reply in the wrong place; I meant it to refer to the item below it (meaning it's not possible to do IO within a value of type Int). You are correct that there must be a function in any program with /at least/ that form (i.e., there could be more statements, but if you want to do processing within pure Haskell, you need to wrap that processing ultimately roughly as shown above). Jon Cast

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popen bug?
by Mark Preston 05 May '03

05 May '03

Is there a bug in the GHC implementation of the popen function that manifests itself when more than 4k is written to the pipe? I'm trying to reproduce in Haskell the standard UNIX shell pipe idiom: cat 4097 | cat where 4097 is a file containing 4097 bytes. My Haskell code looks as follows: -- snip -- import Posix import IO import System main :: IO () main = do sequence_ $ map printChunk chunks printChunk :: String -> IO () printChunk str = do (find_output, _, _) <- popen "cat" [] (Just str) (cat_output, _, _) <- popen "cat" [] (Just find_output) putStrLn cat_output putStrLn $ show (length str) lengths = [4095,4096..] str = repeat 'a' chunks = zipWith take lengths (repeat str) -- end snip -- I get two pieces of output, one for each of the 4095 and 4096 bytes, but nothing for 4097 bytes, the program just hangs. Is this some bizarre feature of lazy evaluation, or a mistake in the way that popen is implemented (or both)? I'm using ghc-5.04.3 on linux 2.4.20-8 (Red Hat 9). Thanks, Mark. _________________________________________________________________ On the move? Get Hotmail on your mobile phone http://www.msn.co.uk/mobile

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RE: Maximum value in a list
by Garner, Robin 05 May '03

05 May '03

You'll find in the Prelude that Int is an instance of Bounded, so minBound and maxBound are defined, and are the identities you need. cheers -----Original Message----- From: sashan [mailto:sashang@ihug.co.nz] Sent: Monday, May 05, 2003 12:30 PM To: haskell-cafe(a)haskell.org Subject: Maximum value in a list Hi I was reading through the tutorial Yet Another Haskell Tutorial and doing chapter 3 ex 5, write a function using foldr and max to return the maximum value in a list. I came up with the following that will work for positive numbers. maxInList :: [Int]->Int maxInList [] = 0 maxInList l = foldr max 0 l Is there an identity for the max function? Because currently if the list [-1,-3,-4] is passed to maxInList it will return 0. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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[Arrays] Random Access Times ?
by Ron de Bruijn 04 May '03

04 May '03

Hi there, I tested below program with for x filled in 1 and 50000. And I saw that when I used 50000 it took more than ten times more time than when I used 1, to complete the expression. So much for randow access memory(RAM). Isn't there somekind of other array that really works with random access? module Test where import IOExts data Lesson = Lesson String Int Int String String deriving Show main = do testing <- newIOArray (0,60000) (Lesson "Hallo" 0 0 "" "") sequence(map(writeIOArray testing x) (test)) a<-readIOArray testing 0 putStr (decompose a) test::[Lesson] test=(replicate 100000 (Lesson "" 1 2 "" "")) decompose (Lesson s1 _ _ _ _) = s1 Greets Ron __________________________________ Do you Yahoo!? The New Yahoo! Search - Faster. Easier. Bingo. http://search.yahoo.com

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RE: Fast Mutable arrays and a general question about IO
by Garner, Robin 04 May '03

04 May '03

Here's an example of mine, that uses a hash table to construct a word frequency table. By no means is this great code - for example the hash table filling up will lead to an infinite loop ... This function produces a word frequency list using an intermediate hash table to accumulate the counts. > makemap :: [String] -> [(String,Int)] > makemap l = filter ((> 0) . snd) $ > runST (emptyHT >>= flip (foldM insertHT) l >>= getElems) The hash table structure, an array.ST > maxHash = 10000 ::Int > > type HT s = STArray s Int (String,Int) > > emptyHT :: ST s (HT s) > emptyHT = newArray (0,maxHash) ("",0) Update the hash table with a single entry > insertHT :: HT s -> String -> ST s (HT s) > insertHT h s = findSlot s h (hash s) >>= updateSlot h s >> return h Finds the correct slot in the hash table, returning the slot index and the old count. > findSlot :: String -> HT s -> Int -> ST s (Int,Int) > findSlot s h i = do (s',c) <- readArray h i > if c == 0 || s == s' > then return (i,c) > else findSlot s h ((i+1) `mod` maxHash) > > updateSlot :: HT s -> String -> (Int,Int) -> ST s () > updateSlot h s (i,n) = writeArray h i (s,n+1) Simple (simplistic ?) hash function > hash :: String -> Int > hash = (`mod` maxHash) . foldl' hash' 0 > where hash' acc c = acc * 2 + ord c

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Re: Fast Mutable arrays and a general question about IO
by Jon Cast 02 May '03

02 May '03

Ron de Bruijn <rondebruijn(a)yahoo.com> wrote: > Hi there, > > I have been busy for a couple of days now, to just get a bit more > feeling on Haskell and I have read a bunch of tutorials, also about > Monads and experimented of course. And now I want to use it for my > real world problem. I think I understand the languageprinciples good > enough, but I yet don't understand the syntax of all of the > functioninterfaces. And that's my problem. > > I want to create the fastest array as possible in Haskell Why? It's not always necessary to go for maximum speed; do you have a reason for needing it in this case? > and I understood that I should use this function(the function below > that creates a mutable array). The only problem is that I don't kwow > how to use it. > > Constructing mutable arrays > > newArray :: (MArray a e m, Ix i) => (i, i) -> e -> m (a i e) > Builds a new array, with every element initialised to undefined. No; that's newArray_. (newArray bounds x) builds an array with every element initialized to x. > Reading and writing mutable arrays > > readArray :: (MArray a e m, Ix i) => a i e -> i -> m e > > Read an element from a mutable array > > writeArray :: (MArray a e m, Ix i) => a i e -> i -> e -> m () > Write an element in a mutable array > > I see it takes two arguments: a tuple of an index type (I use Int, so > for example (0,1) 0 for the lower bound, 1 for the upper bound.), I assume you're talking about newArray here, right? If so, you're correct. > but the e (according to "GHC documentation" it says the "element > type"). Well the type of my elements should be my own datatype > myDataType, but I can't even get it done to create simple mutable > array's of Int or Bool. > How can I make it work with my own datatype myDataType (could you give > some examples for the use of the below functions?)? I think you mean `above'. In any case, could you show us some code you have that's not working? That'll make it much easier for us to answer your questions. > Is it OK to say that in Haskell any value that is put from an extern > source of the program (reading a file, reading stdin etc.) to a really > pure function, must be in a do notation (essentially being a Monadic > operation)? Yes. More specifically, they don't need to use do-notation (although if you don't know why not, don't worry about it), but they do have to be monadic. Specifically, they have to use the IO monad. > So to put it more concrete: In any Haskell program that does IO there > is always some function that has this form: > > do x<-someSource > putBoundVariableToSomeOtherFunctionThatDoesMonadicOperations > (restOfPureFunctionalProgramForExample x) No. > So it's not possible to write some function that does the following: > > f::Int > f do x<-getChar --suppose the Char is a number > someSpecialReturnFunction x > > and than the result of the IO is only Int after some alterations of > above code. Correct. (I don't think this is a big issue; a Haskell program is required to execute in terms of a value `main' in the IO monad.) <snip> Jon Cast

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Re: How to Check whether a file exists and IO[String] vs [String]
by Jon Cast 02 May '03

02 May '03

Alexandre Weffort Thenorio <alethenorio(a)home.se> wrote: > Hello. I am having two problems. I have a program that reads two texts > file and writes a new text with its info. So I have done so that if > main file doesn't exist I just output an error string but my problem > is when it comes to the second file. This file is not exactly needed > so if it is not there, the program should just ignore the file. I > tried doing that but if the second file is there I also get an error > saying that it couldn;t find the file. So is it possible to check > whether a file exists Yes. See Directory.doesFileExist. > or not outputting and empty string if not or the actual data from it > if so without giving an error and making the program not run?? > > For example > > main = do > mainfile <- readFile "xxxxx.txt" > secondfile <- readMyFile "yyyy.txt" > writeFile "test.txt" mainfile++secondfile > > > readMyFile file = do > list <- readFile file > if list == "" then return "" else return list > > Apparently this doesn't work and if file doesn't exist it still gives > me an error. Correct. readFile always assumes the file exists (not to mention, if list == "", then list = "", so we can calculate: readMyFile file = do list <- readFile file if list == "" then return "" else return list = do list <- readFile file if list == "" then return list else return list = do list <- readFile file return list = readFile file so your readMyFile function as written is unnecessary.) If you want to take a different action if the file doesn't exists you have to use doesFileExist as above. > Another problem I ran into is with IO[String] and [String]. > > I have a function like > > foo = do > nf <- readFile "xxx.txt" > return (lines nf) > > Then when I try using foo like a [String] in other programs I run into > type error. True. IO a /= a; they're not even isomorphic. > How can I turn this IO[String] into a [String] so I don't actually > have to open this file in the main and then use it as an argument of > another function running "lines' on main method???? If you want to pass the result into f, you can say: do x <- foo return (f x) or fmap f foo. Jon Cast

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