- Beginners - Haskell.org

types, parentheses, application, composition
by Christopher Howard 26 Nov '12

26 Nov '12

Could someone explain more precisely to me what is the significance of parentheses in the type of an expression? Are they just some kind of syntactic sugar, that simplifies into a type without them, or are they more like parentheses in algebra, or...? In particular, I'm having difficulty understanding the results of composition or application that involves parentheses... I think I've got it figured out and then I find another strange expression that doesn't have the type I expect. Here is an example using the (self-made) function "sqr": code: -------- > :t sqr sqr :: Double -> Double > :t (.) (.) :: (b -> c) -> (a -> b) -> a -> c > :t (. sqr) (. sqr) :: (Double -> c) -> Double -> c > :t ((. sqr) .) ((. sqr) .) :: (a -> Double -> c) -> a -> Double -> c -------- Everything makes sense up until the last expression, "((. sqr) .)". Where did the "(a -> Double -> c)" come from? What's going on here? -- frigidcode.com

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type signature error in a where clause
by Mark Wallace 25 Nov '12

25 Nov '12

I'm writing a merge sort function, but I get type error under such implementation: mergesort :: (a -> a -> Ordering) -> [a] -> [a] mergesort cmp xs = mergeAll (map (\x -> [x]) xs) where mergeAll :: [[a]] -> [a] mergeAll [x] = x mergeAll xs = mergeAll (mergePairs xs) mergePairs :: [[a]] -> [[a]] mergePairs (a:b:xs) = merge a b : mergePairs xs mergePairs xs = xs merge :: [a] -> [a] -> [a] merge as@(a:as') bs@(b:bs') | cmp a b == GT = b : merge as bs' | otherwise = a : merge as' bs merge [] bs = bs merge as [] = as And ghc says: Couldn't match type `a1' with `a' `a1' is a rigid type variable bound by the type signature for merge :: [a1] -> [a1] -> [a1] at /home/ice/Study/Haskell/tutorials/99Questions/21to30.hs:135:7 `a' is a rigid type variable bound by the type signature for mergesort :: (a -> a -> Ordering) -> [a] -> [a] at /home/ice/Study/Haskell/tutorials/99Questions/21to30.hs:124:1 In the first argument of `cmp', namely `a' In the first argument of `(==)', namely `cmp a b' In the expression: cmp a b == GT But if I comment all type signatures, ghc works fine on it. I would really appreciate it if you can point out what causes this question? And how to fix it without changing the structure of the program (i.e. not adding function `cmp' as a parameter of `merge' etc.). Thx.

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Can't install json-.0.4.4 - memory allocation failed
by Dan Tenenbaum 24 Nov '12

24 Nov '12

Hi all, Please tell me where I should post this if this is not the appropriate list. I tried to install the json-0.4.4 package on Mac OS 10.5.8, and got this error: $ sudo cabal install "json-0.4.4" Building json-0.4.4... [1 of 7] Compiling Text.JSON.Types ( Text/JSON/Types.hs, dist/build/Text/JSON/Types.o ) [2 of 7] Compiling Text.JSON.Pretty ( Text/JSON/Pretty.hs, dist/build/Text/JSON/Pretty.o ) [3 of 7] Compiling Text.JSON.ReadP ( Text/JSON/ReadP.hs, dist/build/Text/JSON/ReadP.o ) Text/JSON/ReadP.hs:104:21: Warning: A do-notation statement discarded a result of type b. Suppress this warning by saying "_ <- n", or by using the flag -fno-warn-unused-do-bind [4 of 7] Compiling Text.JSON.Parsec ( Text/JSON/Parsec.hs, dist/build/Text/JSON/Parsec.o ) ghc: memory allocation failed (requested 2097152 bytes) cabal: Error: some packages failed to install: json-0.4.4 failed during the building phase. The exception was: ExitFailure 1 This is on a machine with 16GB of RAM. I don't think http://www.haskell.org/cabal/FAQ.html#cabal-goes-into-an-infinite-loop--run… applies because I'm using higher versions than mentioned there: $ ghc --version The Glorious Glasgow Haskell Compilation System, version 6.12.3 $ cabal --version cabal-install version 0.8.2 using version 1.8.0.6 of the Cabal library I know Mac OS 10.5.8 is old, but unfortunately ugrading this particular machine is not an option. Hope someone can help. Thanks, Dan

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Using StateT with List monad
by Nathan Hüsken 24 Nov '12

24 Nov '12

Hey, I have an example function in the list monad. twoSucc :: Int -> [Int] twoSucc i = [i+1, i+2] Now I want write something similar with the StateT monad transformer. twoSucc :: StateT Int [] () twoSucc = do i <- get put (i+1) -- how do I put [i+1,i+2] here? As mentioned in the comment, when doing runStateT twoSucc n it outputs [((),n+1)] (which makes sense). How do I write it that it behaves similar to the original twoSucc and outputs [((),n+1),((),n+2)]? Thanks! Nathan

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Is all Haskell syntax pointfree compatible?
by Christopher Howard 24 Nov '12

24 Nov '12

After reading the Haskell wiki article about pointfree style I naturally started playing around with with converting my regular code to pointfree. However, I immediately got stuck on the specific cases of 1) do syntax and 2) record syntax. For example: code: -------- playMusic rt = do [source] <- genObjectNames 1 buffer source $= getSound rt "music.wav" sourceGain source $= 0.4 loopingMode source $= Looping play [source] -------- And another (contrived) example: code: -------- data A = A { u :: Int , v :: Double , w :: String , ... } f a b = a { v = b } -------- -- frigidcode.com

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Orphan instances
by Emmanuel Surleau 23 Nov '12

23 Nov '12

Hi there, I have a design issue, which is causing me to generate orphan instances. I'm writing a small program which persists data to a database. In order to be able to switch between different backends, I have a Store typeclass in one module, which is exposed to the frontend, and which has instances in each backend module. But obviously, the backend instances are therefore orphan. Is there a better solution which would let me keep the concrete implementation separate from the interface? Thanks, Emm

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On Windows defragment (Defraggler'ize, etc) the hard drive and turn off your A/V before installing cabal-install :)
by KC 20 Nov '12

20 Nov '12

:) Casey -- -- Regards, KC

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list splitting - nice implementation?
by Emmanuel Touzery 18 Nov '12

18 Nov '12

Hello, i wonder what would be the idiomatic way to achieve that algorithm in haskell: [1,4,56,450,23,46,52] => [1,4,56,450] [1,4,56,450,23,46,52] => [23,46,52] in other words split the list when one element gets smaller than the previous one. Tge rest of the time the list is sorted. There would be only two lists, not N. I always need the first or second sublist, I don't need both at once. But of course a more complete algorithm handling the N case and/or returning both sublists would be good. i could code this by hand, but i'm trying to use as much as possible builtin higher-order functions. However in this case so far I've only come up with this: import Data.List isSorted :: Ord a => [a] -> Bool isSorted l = (sort l) == l secondPart :: Ord a => [a] -> [a] secondPart l = head $ filter isSorted (tails l) firstPart :: Ord a => [a] -> [a] firstPart l = last $ filter isSorted (inits l) It is concise alright, but it seems contrived and also in terms of performance I don't think it's OK (for small lists sure but for big lists?). Anyway, somehow I think something as simple as this must be doable very concisely and with optimal performance using only builtin higher-order functions. Any idea? Thanks! Emmanuel

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Consuming Rule Based Parsing
by Christopher Howard 17 Nov '12

17 Nov '12

I haven't learned anything substantive yet regarding Haskell approaches to parsing, so I would appreciate any pointers towards appropriate modules, abstractions, tutorials, or such like. Specifically: For a particular application, I want to play around with a rules (grammer) based parsing which consumes an incoming data stream. That is, I'm not parsing a single type of document with a set structure. Rather, the idea is that I have a list of symbols (characters, numbers, whatever) and pattern match the front part of the list according to some (grammer) rule (which itself could be a conjunction or disjunction of rules) in a set of rules. If the match succeeds, then I consume that part of the list, and then analyze the remaining list, starting again with the first rule in my rule set. If the match fails, then I try the next rule in my rule set. (Sort of like Prolog's context free grammars and difference lists.) Backtracking would also be cool (to see what other results I can get). Is there a particular Haskell abstraction or system suitable for what I have described? -- frigidcode.com

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Representing most types by the same letter
by Christopher Howard 14 Nov '12

14 Nov '12

Hi. I'm trying to learn Haskore, and it seems rather odd: Why are most types in the system represented by the same letter, the letter 'T'? For example, instead of a Scale type, a Duration type, and a Pitch type, there is a T type in the Scale module, a T type in the Duration module, and a T type in the Pitch module. It makes the documentation rather challenging to read, because I have to read the link for each appearance of the letter to see what type it is referring to. Is there some theory behind this practice? -- frigidcode.com

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