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Implementing C program in Haskell
by Alex Gontcharov 24 Mar '04

24 Mar '04

Hi, Is anyone bored and wants to rewrite a small C program in Haskell?:) http://users.tpg.com.au/galutva/jap.c I am new to Haskell and it'd take me sometime to do it all myself. Thanks

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Where's the error in this snippet of code?
by Alex Gontcharov 24 Mar '04

24 Mar '04

Ignore the layout I can't find the error, running it gives parse error during compile on pStack, it is not very descriptive and I don't what is wrong. --Stack for the digits for numbers, a modulo b digStack :: Integer->Integer->[Integer] digStack a b | a == 0 = [] | otherwise = (mod a b):(digStack (div a b) b) --Takes a list of remainders and pStack :: [Integer] -> Integer -> [Integer] pStack [] _ = [] pStack (x:xs) a = digStack x a ++ pStack xs I think I will never like Haskell :-)

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Re: [Haskell-cafe] Where's the error in this snippet of code?
by Duncan Coutts 24 Mar '04

24 Mar '04

On Wed, 2004-03-24 at 18:00, Alex Gontcharov wrote: > I don't want to hide it within the function, I need the functions > to have file scope. Oh, ok. I assumed from the indentation (you see layout matters!) that it was a nested function. Haskell (like Python) is layout sensitive, the indentation makes a difference. All function definitions in the same scope have to line up in the same column. So for global/file scope functions they should not be indented at all. So the problem with your example was that your second function was indented. Write: globalFunc1 = ... globalFunc2 = ... and not: globalFunc1 = ... globalFunc2 = ... > So, If you want to call a function from another function b I have to > nest it using where? No, only if you want it nested rather than whole file scope. Duncan

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RE: [Haskell-cafe] loop in C , recursion in haskell
by Simon Marlow 24 Mar '04

24 Mar '04

> On Thu, Mar 25, 2004 at 01:38:44AM +1100, Alex Gontcharov wrote: > > Hi, > > > > I am new to haskell and would look to write a function equivalent > > to the following loop in C > > > > int value = 500000; > > int part_stack[4]; > > int *part_ptr = part_stack; > > for (; value; value /= 10000) > > *part_ptr++ = value % 10000; > > part_stack :: [Int] > part_stack = [0,50] > > Note that I've performed a memoization optimization--this > makes the code both smaller, faster and easier to read! :P Ignore David, he's pulling your leg. Here's the proper translation: do alloca $ \value -> do poke value (500000::Int) allocaArray 4 $ \part_stack -> do alloca $ \part_ptr -> do poke part_ptr part_stack let loop = do val <- peek value if val == 0 then return () else do p <- peek part_ptr poke p (val `rem` 10000) poke part_ptr (p `plusPtr` 1) poke value (val `quot` 10000) loop loop Cheers, Simon

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loop in C , recursion in haskell
by Alex Gontcharov 24 Mar '04

24 Mar '04

Hi, I am new to haskell and would look to write a function equivalent to the following loop in C int value = 500000; int part_stack[4]; int *part_ptr = part_stack; for (; value; value /= 10000) *part_ptr++ = value % 10000; Thanks

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Re: [Haskell-cafe] What are Kind errors and how do you fix them?
by MR K P SCHUPKE 24 Mar '04

24 Mar '04

Erm... okay my mistake thats not the problem... although it is to with state... When decaring a monad the return type is included in the class definition (it is a constructor class) This means the return value MUST be the last parameter to the type, so you need: type ReverseType a string = (string ->(string,a)) data Reverse string a = Reverse (ReverseType a string) instance Monad (Reverse s) where return x = Reverse (\text -> (text,x)) (Reverse p) >>= k = Reverse p3 where p3 s0 = p2 s1 where (Reverse p2) = k a (s1,a)=p s0 I was confused by the unusual format for bind... I would wtite this: instance Monad (Reverse s) where return x = Reverse $ \s -> (s,x) (Reverse p) >>= k = Reverse $ \s -> case p s of (s',a) -> (\(Reverse r) -> r) (k a) s' or even better define the data type as: data Reverse string a = Reverse { runReverse :: ReverseType a string } instance Monad (Reverse s) where return x = Reverse $ \s -> (s,x) (Reverse p) >>= k = Reverse $ \s -> case p s of (s',a) -> runReverse (k a) s' Regards, Keean.

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Re: [Haskell-cafe] What are Kind errors and how do you fix them?
by MR K P SCHUPKE 24 Mar '04

24 Mar '04

I dont understand what you are trying to do: type ReverseType a string = (string ->(string,a)) takes a string an returns a string and another value (looks like the state monad with state=String. data Reverse a string = Reverse (ReverseType a string) this just declares a type wrapper... why have both of the above, you only need: data Reverse a s = Reverse (s -> (s,a)) NOw for your class definition: return x = Reverse (\text -> (text,x)) this is the standard return from a state monad. (Reverse p) >>= k = Reverse p3 where p3 s0 = p2 s1 where (Reverse p2) = k a (s1,a)=p s0 the second parameter (k) needs the state passed as well after unwrapping like: p2 = (\(Reverse x) -> x) (k a) s1 Regards, Keean.

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Re: [Haskell-cafe] class Runnable vs fromJust (was Re: haskell idiom for reversible computations)
by MR K P SCHUPKE 23 Mar '04

23 Mar '04

The problem with Runnable is that it is not ewasily implementable for all monads... So purists rather than implement it for some don't implement it at all. Here's some examples: class Runnable x y where run :: x -> y instance Runnable (m a) (m a) where run = id instance Runnable (s -> m a) (s -> m a) where run = id instance (Monad m,Monad n,MonadT t m,Runnable (m a) (n a)) => Runnable (t m a) (n a) where run = run . down instance (Monad m,MonadT t m,Monad (t m)) => Runnable (t m a) (m a) where run = down These declare runnable for monad-transformers for which "down" is definable down is the functional opposite of lift (or "up" as it is sometimes called) The state monad transformer needs an instance of Runnable if you want to pass a value in: instance (MonadState st (StateT st m),Monad m,Monad n,Runnable (st -> m s) (st -> n s)) => Runnable (StateT st m s) (st -> n s) where run = run . (\(ST m) s -> do (_,a) <- m s return a) instance (MonadState st (StateT st m),Monad m) => Runnable (StateT st m s) (st -> m s) where run = \(ST m) s -> do (_,a) <- m s return a Whereas defining Runnable for the continuation-monad-transformer is quite a challenge: instance (MonadT (ContT r) m,Runnable ((r -> m r) -> m r) ((r -> n r) -> n r)) => Runnable (ContT r m r) ((r -> n r) -> n r) where run = run . (\(CT m) kappa -> m kappa) instance MonadT (ContT r) m => Runnable (ContT r m r) ((r -> m r) -> m r) where run = (\(CT m) kappa -> m kappa) Have Fun. Keean.

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asking for a favor
by garroyo＠dsic.upv.es 23 Mar '04

23 Mar '04

Hi, Does someone has the Haskell XML Toolbox, lets say working in Windows environment?. I have tried install them in Hugs 98 for Windows but there are Hexadecimal characters escape out of range. One choice, I know, is go to LINUX. I am working in CNC programming and I am in the begining of the a research. Thanks in advace. Gustavo. ------------------------------------------------- This mail sent through IMP: http://horde.org/imp/

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haskell idiom for reversible computations
by S. Alexander Jacobson 23 Mar '04

23 Mar '04

If I am writing a rener/parser for e.g. HTTP Requests and Responses, I would like to be able to guarantee that the parser really is the inverse of the generator. I assume that if I define a pair of computations as reversible then I can put those computations in some monad such that I can do: parse = inverse generate and parse2 = inverse do {generate;generate2} Actually, I assume this is an arrow rather than a monad because a render is a computation that takes a value and produces a value rather than simply a computation that produces a value. Whether its an arrow or not, I also assume this problem is sufficiently common that there is a well worked out idiom for this such that I don't have to hand-code and prove inverse functions for all my data structures. I just don't know what it is. -Alex- _________________________________________________________________ S. Alexander Jacobson mailto:me@alexjacobson.com tel:917-770-6565 http://alexjacobson.com

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