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Re: Using "type"
by Ashley Yakeley 05 May '02

05 May '02

At 2002-05-04 12:45, Samuel E. Moelius III wrote: > type Thing m = m () > > type Const a b = a > > f :: Thing m -> Thing m > f x = x > > test :: Thing (Const Int) -> Thing (Const Int) This is wrong. Type synonyms must be fully applied, whereas you have Const applied only to one argument, not two. 'Const' isn't really a type-constructor, it's more like a macro. Is it something in the air? Spring is here, and everyone wants type-lambda... -- Ashley Yakeley, Seattle WA

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Using "type"
by Samuel E. Moelius III 05 May '02

05 May '02

I apologize if this question has been asked before... It seems to me the following code should be legal: type Thing m = m () type Const a b = a f :: Thing m -> Thing m f x = x test :: Thing (Const Int) -> Thing (Const Int) test = f However, GHC gives the following error: Couldn't match `Thing m' against `Thing (Const Int)' Expected type: Thing m Inferred type: Thing (Const Int) Expected type: Thing (Const Int) -> Thing (Const Int) Inferred type: Thing m -> Thing m But if you change the definition of Const newtype Const a b = MakeConst a the code compiles fine. Is this the desired behavior? Is it the case that when you have quantification of a type variable of kind * -> * (in this case, m), that variable may only become bound to a type constructor? Sam Moelius

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Re: tricky instance declarations
by Ashley Yakeley 03 May '02

03 May '02

At 2002-05-03 03:53, Warrick Gray wrote: >instance Monad (IO (Either ConnError a)) where You do of course mean instance Monad (/\a => IO (Either ConnError a)) where ... ...but you just can't do type-lambdas in Haskell. -- Ashley Yakeley, Seattle WA

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User-Defined Types
by Matthias 02 May '02

02 May '02

Hello, I am a student of computer science at the university of Aachen in Germany and presently do a course about functional programming in haskell. I just wanted to introduce myself, because it is likely to happen that my posting apear frequently in the next couple of month. So my first question is about User-Defined Types. Stating the Gentle Introduction to Haskell 98 (2.2) , i already know that: "Type Constructors and date Constructors are in seperte namespaces. [example]" But according to the chapter (2.2.1) "Recursive Types" i see the polymorphic definition of a tree is: data Tree a = Leaf a | Branch (Tree a) (Tree a) where (Tree a) is obviously the data-constructor out of "the other namespace". I guess that the () is warping Tree a from the "data-constructor-namespace" into the "type-constructor-namespace". Since the gentle-intro implicitly uses (), which i am not used to - i would be happy about some hints about (). That might not be to difficult, i hope. Also i would not regret some hints about "->" which is used in function-type-definitions. I would prefer writing plus :: a,a -> a instead of the ->-variant in order to stress the distinction between arguments and returned type. What is meant that -> is "associating to the right"? Is there a name for '->'? Sincerely Matthias Pfeifer

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Re: Read File
by Jorge Adriano 01 May '02

01 May '02

----------------- [moving to haskell-cafe] > I want to read some data from a text file. I accomplished it but I want to > read data from file as a Integer list for example text file has [1,2,3,5] ----------------- Wait you actually have "[1,2,3,5]" --------------------- > and when I read this data from file I handeled it as string. What can I do > to get it as integer list. Hint: map read ["1","2","3"] ::[Int] ------------------- this would be helpfull if you had "1 2 3 5". It's easier then Hint: read "[1,2,3,5]"::[Int] J.A.

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Re: Read File
by Jorge Adriano 01 May '02

01 May '02

[moving to haskell-cafe] > I want to read some data from a text file. I accomplished it but I want to > read data from file as a Integer list for example text file has [1,2,3,5] > and when I read this data from file I handeled it as string. What can I do > to get it as integer list. Hint: map read ["1","2","3"] ::[Int] J.A.

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(no subject)
by Ricky Kurniawan 01 May '02

01 May '02

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some algorithms are hard to code in O(n) in Haskell
by klusacek＠atrey.karlin.mff.cuni.cz 29 Apr '02

29 Apr '02

Hi - does anybody know how to do a term unification in Haskell in O(n), where n is length of the input terms? I know Montanari-Martelli algorithm for this, but I'm unable to code it in Haskell (in O(n)). I would need a destructive update of the directed graph representing terms (to do a substitution) and checking for cycles in directed graphs (these cycles emerge when unifying things like x=f(x) -- their detection is also called occurs-check). I've been trying hard but still have no solution. Maybe I'm not smart enough, maybe I don't know some trick, maybe it's impossible in Haskell. Thanks, _dave_

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Re: Binary search tree
by Rijk J. C. van Haaften 29 Apr '02

29 Apr '02

>Hi everybody, > >I studied haskell this semester at the university and I was required to >implement a binary search tree in haskell. >I would appreciate if anyone could send me an example code of this data >structure. Just read a standard textbook. Some useful course notes by Jeroen Fokker are available online at http://www.cs.uu.nl/people/jeroen/courses/fp-eng.pdf Enough to solve your problem. Rijk-Jan van Haaften

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Binary search tree
by Claudio Marcio de Souza Vicente 29 Apr '02

29 Apr '02

Hi everybody, I studied haskell this semester at the university and I was required to implement a binary search tree in haskell. I would appreciate if anyone could send me an example code of this data structure. thanks in advance, Claudio

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