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Question about implementing an off-side rule in Parsec 2
by Bas van Gijzel 29 Apr '09

29 Apr '09

Hello everyone, I'm doing a bachelor project focused on comparing parsers. One of the parser libraries I'm using is Parsec (2) and I'm going to implement a very small subset of haskell with it, with as most important feature the off-side rule (indentation based parsing) used in function definitions and possibly in where clauses. But I'm still a bit stuck on how to implement this cleanly. I tried to search for some examples on blogs but I haven't found anything yet. As far as I can see the way to go would be using getState and updateState methods defined in Parsec.Prim and to use the methods in Parsec.Pos to compare the difference in indendation for tokens. But I haven't completely wrapped my head around any state monad yet and I don't understand Parsec enough yet to see how to use the methods Parsec.Pos and state easily. Some examples or pointers to something to read would really be helpful. Thanks in advance, Bas van Gijzel

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Re: [Haskell-cafe] chr/ord?
by michael rice 29 Apr '09

29 Apr '09

Thank guys, Now what am I misunderstanding in the code below? I would think that *Main> comb (Just 65) foo and *Main> comb (lookup 'A' lst) foo would return the same result Just 'A' Michael ===========Haskell code============= import Data.Char lst = [('A',65),('B',66),('C',67),('D',68)] -- comb is a combinator for sequencing operations that return Maybe comb :: Maybe a -> (a -> Maybe b) -> Maybe b comb Nothing _ = Nothing comb (Just x) f = f x foo :: Int -> Maybe Char foo n = Just (chr n) ================Ghci=========== [michael@localhost ~]$ ghci GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Prelude> :load test1 [1 of 1] Compiling Main ( test1.hs, interpreted ) Ok, modules loaded: Main. *Main> comb (Just 65) foo Just 'A' *Main> comb Nothing foo Nothing *Main> lookup 'A' lst Just 65 *Main> (lookup 'A' lst) Just 65 *Main> comb (lookup 'A' lst) foo <interactive>:1:22: Couldn't match expected type `Integer' against inferred type `Int' In the second argument of `comb', namely `foo' In the expression: comb (lookup 'A' lst) foo In the definition of `it': it = comb (lookup 'A' lst) foo *Main> --- On Tue, 4/28/09, Lennart Augustsson <lennart.augustsson(a)gmail.com> wrote: From: Lennart Augustsson <lennart.augustsson(a)gmail.com> Subject: Re: [Haskell-cafe] chr/ord? To: "Tim Wawrzynczak" <inforichland(a)gmail.com> Cc: "michael rice" <nowgate(a)yahoo.com>, "haskell-cafe(a)haskell.org" <haskell-cafe(a)haskell.org> Date: Tuesday, April 28, 2009, 10:53 PM You can also use the more general toEnum and fromEnum. -- Lennart (iPhone) On Apr 29, 2009, at 2:13, Tim Wawrzynczak <inforichland(a)gmail.com> wrote: Michael, those functions are not in the Prelude, they're in Data.Char. On Tue, Apr 28, 2009 at 8:08 PM, michael rice <nowgate(a)yahoo.com> wrote: Hi, My Prelude docs must be out of date because chr and ord don't seem to be there. How do I access these functions? Michael =============== [michael@localhost ~]$ ghci GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Prelude> chr 65 <interactive>:1:0: Not in scope: `chr' Prelude> ord 'A' <interactive>:1:0: Not in scope: `ord' Prelude> _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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name for monad-like structure?
by Michael Vanier 29 Apr '09

29 Apr '09

I've stumbled upon a structure that is like a weaker version of a monad, one that supports return and >> but not >>=. Has anyone seen this before, and if so, does it have a standard name? Mike

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breaking too long lines
by Christian Maeder 28 Apr '09

28 Apr '09

Hi, according the several style guides, lines shouldn't be too long (longer than 78 characters). http://www.cs.caltech.edu/courses/cs11/material/haskell/misc/haskell_style_… http://www.haskell.org/haskellwiki/Programming_guidelines However, I miss hints how to break lines best. Therefore I make some suggestions here and ask for comments. If a "one-liner" does not fit on one line I break the line after "=" rather than breaking the following infix- or prefix-expression. (I break in the same way after "->" or "<-" in case- or do- expressions) (Some people move "=" to the next line, but I only suggest this for proper infix operators, below.) If a "do" or a short "case ... of" or "let ... in" fits, I leave it at the end of the previous line behind "=" ("->" or "<-"). A line should be broken after "do" or "of" (from "case") in order to allow insertions without breaking the layout, provided there is preceding text at all. The following is fine, because there's no text before "do") c x = do y <-... z ... But this is fine, too, if not better: c x = do y <- ... z ... It's not necessary to put "do" (or "let") on a separate line (and care about the indentation of that keyword): c x = do y <-... z ... Because layout may easily break (if b is renamed) and the layout block starts already too far to the right, this is really bad: c b x = unless b $ do y <- ... z ... In many cases a "do" can be moved to the end of the previous line: c b x = unless b $ do y <- ... z What applies to "do" could be applied to "let" as well (and vice versa), but: f x = let y = x + x z = y + y in z does not look as nice as: f x = let y = x + x z = y + y in z And also a "let" (without "in") within a "do" should not be broken after "let". A long infix-expression should be broken _before_ an infix symbol to better indicate it's continuation: f ++ g ++ h f . g $ h x (One should also put spaces around infix operators and should not put unnecessary brackets around prefix applications.) Surely a long prefix expression can be broken anywhere, but I try to break expressions on the top-level and not within too deeply nested sub-expressions: f arg1 arg2 arg3 (longArg4 sa1 sa2 sa3 sa4) arg5 arg6 arg7 After a line break (following "=", "do", "<-", or "->") I try to stay as far to the left as layout permits (using 2 spaces as minimal indentation). I don't care if "=", <-", or "->" of one block line up (which surely is difficult for the top-level module block) and I may put parts of the rhs below parts of the lhs: let longPattern = case bla of Nothing -> "don't know" Just b -> "a longer expression" vN = ... The standard breaking of if-then-else (within "do") is: if ... then ... else ... but I think the following variations are also fine: ... <- if ... then ... else ... if ... then ... else ... By chance I rarely use guards and "where", therefore I give no examples for those expressions. Somewhat tricky I've found are 2 do-expressions connected by an infix-operator within an outer do-expression: do c <- letter do d <- digit return [c, d] <|> do u <- char '_' return [c, u] The line containing "<|> do" is critical to indentation. If "<|> do" is moved (2 columns) to the left it'll be wrong (by chance c will not be in scope). If "<|> do" is moved (3 columns) to the right it'll mean something different, namely parsing and returning a letter and a digit _or_ parsing a letter, a digit, and an underscore, but only return the letter and the underscore. In order to more clearly put the infix operator into the middle I've tried to insert one more space: do c <- letter do d <- digit return [c, d] <|> do u <- char '_' return [c, u] Also the indentation of data types is an issue, because I think, code shouldn't be indented due to long (constructor) names. I've nothing against long names, but one shouldn't try to put blocks to the right of them: data TName = LongConstructorName { selector1 :: C1 , ... } | LongSecondConstructor .... deriving ... Cheers Christian

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Is curryfication practical with dynamic scope?
by Loup Vaillant 28 Apr '09

28 Apr '09

Hi, can this code (\a -> \b -> a) 1 2 eval to normal form in a strict language with dynamic scope? Thanks, Loup PS: some context: I am currently implementing a toy language. The AST is this one: data whnf = -- Weak Head Normal Form | Unit | Int of int | Bool of bool | Prim of (whnf -> whnf) -- curried | Fun of string * expr -- curried data expr = | Whnf of whnf | Var of string | App of expr * expr | If of expr * expr * expr Dynamic typing, dynamic scope, and, as you can see, functions have to be curried. I plan to implement lexical scope.

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Deriving type family data
by Tuve Nordius 28 Apr '09

28 Apr '09

If I for some data type wants to derive, in this case Data and Typeable for use with syb code, but the problem is the same regardless what I want to derive. data family Something data Tree = Leaf Something | Fork Something Tree Tree deriving (Data, Typeable) The problem is I want to derive a class for a data type that depends on some non instantiated data type. I can of course rewrite Tree as: data (Data a) => Tree a = Leaf a | Fork a Tree Tree deriving (Data, Typeable) but then data family Something is redundant, I want to think of Something as a not yet instantiated abstract data type Is there anyway to express that either the instance of Something used in Tree should be a member of Data, or even better any instance of Something should be a member of Data or maybe even better that every instance should derive data. data family Something deriving Data //Tuve Nordius

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TraverseAccum: an effectful accumulating map.
by Florent BALESTRIERI 28 Apr '09

28 Apr '09

Hello fellow haskellers. This message is a valid literate haskell file. I ran it with ghc 6.10.1 > module TraverseAccum where > import Control.Applicative > import Data.Traversable A year ago, I read "Why Attribute Grammars Matter". In it we found a function on lists wich combined three traversals into one to compute the differences to the average: > avg_diff_list :: [Double] -> [Double] > avg_diff_list xs = > let nil = const (0,0.0,[]) > cons x f mean = > let (n,s,ds) = f mean > in (n+1,s+x, (x - mean) : ds) > (n,s,ds) = foldr cons nil xs (s/n) > in ds Then I found "Applicative programming with effects" which abstracted the notions of traversal and folding. We can try and generalize avg_diff_list to work over any Traversable data structure. Here is a possible version with two traversals: > avg_diff' :: (Traversable t, Fractional f) > => t f -> t f > avg_diff' t = > let start = (0,0.0) > step (n,s) x = ((n+1, s+x), \mean -> x - mean) > ((n,s), diffs) = mapAccumL step start t > in sequenceA diffs (s/n) But we can do better with only a single traversal. The idea is to define a function traverseAccumL which does the work of mapAccumL but with an effectful map. (mapM) The definition of mapAccumL uses a StateL applicative functor whereas traverseAccumL uses a StateL monad transformer. First a few definitions. > -- left-to-right state transformer > newtype StateLT m s a = StateLT { runStateLT :: s -> m (s, a) } > instance Functor m => Functor (StateLT m s) where > fmap f (StateLT k) = StateLT $ \ s -> fmap f <$> k s > instance (Functor m, Monad m) => Applicative (StateLT m s) where > pure x = StateLT (\ s -> return (s, x)) > StateLT kf <*> StateLT kv = StateLT kfv > where kfv s = do (s', f) <- kf s > (s'', v) <- kv s' > return (s'', f v) Incidently, I always wondered why Monad wasn't a subclass of Functor. > traverseAccumL :: ( Traversable t, Functor m, Monad m) > => (a -> b -> m (a, c)) -> a -> t b -> m (a, t c) > traverseAccumL f s t = runStateLT (traverse (StateLT . flip f) t) s I could have done without StateLT but the signature would have been different: traverseAccumL :: (Traversable t, Monad m) => (a -> b -> m (c, a)) -> a -> t b -> m (t c, a) traverseAccumL f s t = runStateT (mapM (StateT . flip f) t) s Now the definition of avg_diff calls traverseAccumL with the reader monad > avg_diff :: (Traversable t, Fractional f) > => t f -> t f > avg_diff t = > let start = (0,0.0) > step (n,s) x mean = ((n+1, s+x), x - mean) > ((n,s), diffs) = traverseAccumL step start t (s/n) > in diffs The dual definition traverseAccumR is almost the same. The only difference lies in the definition of <*> StateRT kf <*> StateRT kv = StateRT kfv where kfv s = do (s', v) <- kv s (s'', f) <- kf s' return (s'', f v) > -- right-to-left state transformer > newtype StateRT m s a = StateRT { runStateRT :: s -> m (s, a) } > > instance Functor f => Functor (StateRT f s) where > fmap f (StateRT k) = StateRT $ \ s -> fmap f <$> k s > > instance (Functor m, Monad m) => Applicative (StateRT m s) where > pure x = StateRT (\ s -> return (s, x)) > StateRT kf <*> StateRT kv = StateRT kfv > where kfv s = do (s', v) <- kv s > (s'', f) <- kf s' > return (s'', f v) > > traverseAccumR :: ( Traversable t, Functor m, Monad m) > => (a -> b -> m (a, c)) -> a -> t b -> m (a, t c) > traverseAccumR f s t = runStateRT (traverse (StateRT . flip f) t) s -- Florent B.

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Can subclass override its super-class' default implementation of a function?
by siki 28 Apr '09

28 Apr '09

I'm not sure if this is possible at all. I'd like to do something like this: class A a where foo :: a -> Double foo a = 5.0 class (A a) => B a where foo a = 7.0 data Blah = Blah data Bar = Bar instance A Blah instance B Bar let blah = Blah bar = Bar foo blah -- should print out 5 foo bar -- should print out 7 Basically, I have a bunch of instances that have a common functionality but I'd like to be able to group those instances and give each group a different default implementation of that functionality. It's so easy to do this in Java, for example, but I have no idea how to do it in Haskell. The above example will not compile, btw. Thanks a lot -- View this message in context: http://www.nabble.com/Can-subclass-override-its-super-class%27-default-impl… Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

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Re: [Haskell-cafe] Random number example
by michael rice 28 Apr '09

28 Apr '09

How do I define type Random [Int] for rollNDice in Exercise 1, given the code below? Michael ============ Exercises 1. Implement rollNDice :: Int -> Random [Int] from the previous subsection with >>= and return. NOTE: Since >>= and return are already present in the Prelude, you may want to use import Prelude hiding ((>>=),return) to avoid compilation errors. ================= {-# LANGUAGE NoImplicitPrelude #-} import Prelude hiding ((>>), (>>=), return) type Seed = Int type Random a = Seed -> (a, Seed) randomNext :: Seed -> Seed randomNext rand = if newRand > 0 then newRand else newRand + 2147483647 where newRand = 16807 * lo - 2836 * hi (hi,lo) = rand `divMod` 127773 rollDie :: Random Int rollDie seed = ((seed `mod` 6) + 1, randomNext seed) (>>) :: Random a -> Random b -> Random b (>>) m n = \seed0 -> let (result1, seed1) = m seed0 (result2, seed2) = n seed1 in (result2, seed2) (>>=) :: Random a -> (a -> Random b) -> Random b (>>=) m g = \seed0 -> let (result1, seed1) = m seed0 (result2, seed2) = (g result1) seed1 in (result2, seed2) return :: a -> Random a return x = \seed0 -> (x, seed0) sumTwoDice :: Random Int sumTwoDice = rollDie >>= (\die1 -> rollDie >>= (\die2 -> return (die1 + die2))) rollNDice :: Int -> Random [Int] --- On Thu, 4/23/09, michael rice <nowgate(a)yahoo.com> wrote: From: michael rice <nowgate(a)yahoo.com> Subject: Re: [Haskell-cafe] Random number example To: "Ross Mellgren" <rmm-haskell(a)z.odi.ac> Cc: haskell-cafe(a)haskell.org Date: Thursday, April 23, 2009, 5:49 PM Hi Ross, Thanks for going the extra mile. A lot of what you did I haven't seen before, so it's going to take me some time to go through it. But I'll be back. Michael --- On Thu, 4/23/09, Ross Mellgren <rmm-haskell(a)z.odi.ac> wrote: From: Ross Mellgren <rmm-haskell(a)z.odi.ac> Subject: Re: [Haskell-cafe] Random number example To: "michael rice" <nowgate(a)yahoo.com> Cc: haskell-cafe(a)haskell.org Date: Thursday, April 23, 2009, 11:51 AM So there are a couple problems. First is you are trying to rebind prelude functions, when instead you should be creating an instance of Monad. This requires a bit of shuffling because without language extensions you can't instance Monad Random for your type of Random, as it is a type synonym. So, changing the type synonym to a newtype and instancing monad, you get: module Rand9b where import Control.Applicative (Applicative(..), (<$>), (<*>))import Control.Monad (ap, liftM) type Seed = Intnewtype Random a = Rand { unRand :: (Seed -> (a, Seed)) } randomNext :: Seed -> SeedrandomNext rand = if newRand > 0 then newRand else newRand + 2147483647 where newRand = 16807 * lo - 2836 * hi (hi,lo) = rand `divMod` 127773 rollDie :: Random IntrollDie = Rand $ \ seed -> ((seed `mod` 6) + 1, randomNext seed) instance Monad Random where (>>=) = randomBind return = randomReturn instance Functor Random where fmap = liftM instance Applicative Random where pure = return (<*>) = ap randomBind :: Random a -> (a -> Random b) -> Random bm `randomBind` g = Rand $ \seed0 -> let (result1, seed1) = unRand m $ seed0 (result2, seed2) = unRand (g result1) $ seed1 in (result2, seed2) randomReturn :: a -> Random arandomReturn x = Rand $ \ seed0 -> (x, seed0) sumTwoDice :: Random IntsumTwoDice = (+) <$> rollDie <*> rollDie I also threw in instances of Functor and Applicative, so that I could simplify sumTwoDice using applicative form (much nicer, no? Applicative is totally rockin') Now you need one more thing, a way to convert a series of Random actions into a pure function: runRandom :: Seed -> Random a -> arunRandom s f = fst . unRand f $ s which now makes what you want to do in GHCi easy and well wrapped: Prelude> :reload[1 of 1] Compiling Rand9b ( rand9b.hs, interpreted )Ok, modules loaded: Rand9b.*Rand9b> runRandom 0 sumTwoDice3 Hope this helps,-Ross On Apr 23, 2009, at 11:28 AM, michael rice wrote: I pretty much followed the sequence of steps that led to this final code (see below), but will be looking it over for a while to make sure it sinks in. In the meantime, I get this when I try to use it (sumTwoDice) at the command line: [michael@localhost ~]$ ghci rand9 GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main ( rand9.hs, interpreted ) Ok, modules loaded: Main. *Main> sumTwoDice <interactive>:1:0: No instance for (Show (Seed -> (Int, Seed))) arising from a use of `print' at <interactive>:1:0-9 Possible fix: add an instance declaration for (Show (Seed -> (Int, Seed))) In a stmt of a 'do' expression: print it *Main> Can I employ a 'do' expression from the command line? Also, can I now use functions (>>) (>>=) and 'return' defined in the Prelude and still have this code work? Michael ================== {-# LANGUAGE NoImplicitPrelude #-} import Prelude hiding ((>>), (>>=), return) type Seed = Int type Random a = Seed -> (a, Seed) randomNext :: Seed -> Seed randomNext rand = if newRand > 0 then newRand else newRand + 2147483647 where newRand = 16807 * lo - 2836 * hi (hi,lo) = rand `divMod` 127773 rollDie :: Random Int rollDie seed = ((seed `mod` 6) + 1, randomNext seed) (>>) :: Random a -> Random b -> Random b (>>) m n = \seed0 -> let (result1, seed1) = m seed0 (result2, seed2) = n seed1 in (result2, seed2) (>>=) :: Random a -> (a -> Random b) -> Random b (>>=) m g = \seed0 -> let (result1, seed1) = m seed0 (result2, seed2) = (g result1) seed1 in (result2, seed2) return :: a -> Random a return x = \seed0 -> (x, seed0) sumTwoDice :: Random Int sumTwoDice = rollDie >>= (\die1 -> rollDie >>= (\die2 -> return (die1 + die2))) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -----Inline Attachment Follows----- _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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Converting IO [XmlTree] to [XmlTree]
by rodrigo.bonifacio 28 Apr '09

28 Apr '09

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