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HLint: use void instead of return()
by harry 22 Apr '13

22 Apr '13

I have some monadiccode of the form: foo >>= something >>= bar >> return () HLint is telling me to change this to: void (foo >>= something >>= bar) Why is the second form considered better? I actually find the original to be clearer.

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Monad instances and type synonyms
by Christopher Howard 16 Apr '13

16 Apr '13

I am playing around with some trivial code (for learning purposes) I wanted to take code: -------- -- SaleVariables a concrete type defined early -- `Adjustment' represents adjustment in a price calculation -- Allows functions of type (a -> Adjustment a) to be composed -- with an appropriate composition function type Adjustment a = SaleVariables -> a -------- And put it into code: -------- instance Monad Adjustment where (>>=) = ... return = ... -------- If I try this, I get code: -------- Type synonym `Adjustment' should have 1 argument, but has been given none In the instance declaration for `Monad Adjustment' -------- But if I give an argument, then it doesn't compile either (it becomes a "*" kind). And I didn't want to make the type with a regular "data" declaration either, because then I have to give it a constructor, which doesn't fit with what I want the type to do. -- frigidcode.com

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Down in Data.Ord vs GHC.Exts
by Matthew O'Connor 16 Apr '13

16 Apr '13

While attempting to do some sorting, I ran across Down. The docs: http://hackage.haskell.org/packages/archive/base/latest/doc/html/Data-Ord.h… it's in Data.Ord, but GHCi 7.4.1 complains the Down data constructor isn't in scope, but if I import GHC.Exts ( http://hackage.haskell.org/packages/archive/base/latest/doc/html/GHC-Exts.h…) I get the constructor I expect. Are the docs wrong? Or is there something I'm not doing right to get Down exported from Data.Ord? Or other things to check? Thanks, Matthew

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Re: [Haskell-beginners] help with lists and tuples
by Miro Karpis 15 Apr '13

15 Apr '13

millions of thanks for all the info. m. On Mon, Apr 15, 2013 at 1:21 AM, Paul Higham <polygame(a)me.com> wrote: > Wouldn't it do just to have > > nActivationSum = map nActivation > > or am I missing your intent here? > > Note that on the left-hand side of the last equation inputs would have to > have type (Float, Float) and weights would have to have type > [(Float,Float)], but on the right-hand side both inputs and weights have to > be Float. The result of > nActivation (inputs, weights) is Float, but the result of > nActivationSum (inputs:weights) should be [(Float,Float)], hence the error > message. > > :: paul > > On 2013-04-14, at 16:04 , Miro Karpis wrote: > > Please can you help me with this? What I want is to have on input a list > of tuple (size 2) and on output list of floats. So now I'm returning only > float instead of a list. But unfortunately I can't figure out how to > implement the list...... > > > sigmoid :: Float -> Float > sigmoid x = 1.0 / (1 + exp (-x)) > > nActivation :: (Float,Float) -> Float > nActivation (x,y) = sigmoid(x*y) > > nActivationSum :: [(Float, Float)] -> [Float] > nActivationSum [] = [] > nActivationSum (inputs:weights) = *nActivation* (inputs, weights) > > > > Couldn't match expected type `[Float]' with actual type `Float' > In the return type of a call of `nActivation' > In the expression: nActivation (inputs, weights) > In an equation for `nActivationSum': > nActivationSum (inputs : weights) = nActivation (inputs, weights) > > > thanks, > m. > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > >

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help with lists and tuples
by Miro Karpis 15 Apr '13

15 Apr '13

Please can you help me with this? What I want is to have on input a list of tuple (size 2) and on output list of floats. So now I'm returning only float instead of a list. But unfortunately I can't figure out how to implement the list...... sigmoid :: Float -> Float sigmoid x = 1.0 / (1 + exp (-x)) nActivation :: (Float,Float) -> Float nActivation (x,y) = sigmoid(x*y) nActivationSum :: [(Float, Float)] -> [Float] nActivationSum [] = [] nActivationSum (inputs:weights) = *nActivation* (inputs, weights) Couldn't match expected type `[Float]' with actual type `Float' In the return type of a call of `nActivation' In the expression: nActivation (inputs, weights) In an equation for `nActivationSum': nActivationSum (inputs : weights) = nActivation (inputs, weights) thanks, m.

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Will Haskell Platform 2012.4.0.0 uninstall cleanly for the installation of 2013.2.0.0
by KC 13 Apr '13

13 Apr '13

:) -- -- Regards, KC

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Warsaw Haskell Group
by Przemyslaw Kaminski 13 Apr '13

13 Apr '13

Hello! We are organizing a group of Haskell fans from Warsaw, Poland. So far we have set up a community on G+ (https://plus.google.com/u/0/communities/103183708602453146804) a mailing list is on the way. We would like to meet regularly someplace in the capital (we still decide where), and hack on Haskell together as much as we can :) Should there be more people involved, we could arrange regular lectures. Hope to see you there! Warsaw Haskell Group

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FAQ: When to ask on Haskell-Beginners, when on Haskell-Cafe (was haskell testing framework - missing htfpp)
by Kim-Ee Yeoh 13 Apr '13

13 Apr '13

On Thu, Apr 11, 2013 at 10:50 PM, Chaddaï Fouché <chaddai.fouche(a)gmail.com> wrote: > More probably you should rather have asked this on Haskell-Cafe, it's not > really a beginner question. htfpp sounds like a pre processor The short answer is: if your question falls within the scope of LYAH, then it belongs here. If it's package specific (wxhaskell, htfpp), you're probably better off on the bigger list. Understandably, there's a huge gap between the two. The rule of thumb I'd use (and this is certainly not without controversy) is: when in doubt use Cafe. -- Kim-Ee

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Multi-param type classes and type dependency
by Olivier Iffrig 12 Apr '13

12 Apr '13

Hello, I'm trying to write some code which boils down to this idea : I have a collection (Collection), and I want to play with some annotations (of any type) to the items of the collection. I will often have only one collection and many different annotation sets. What matters to me is to be able to have an arbitrary index type (or at least instances of Ix) for the collection, and use it with the annotations as well. I've come to this code so far : -------------------- {-# LANGUAGE MultiParamTypeClasses, FlexibleInstances #-} -- Just for the example data Thing = Foo | Bar deriving Show -- A collection -- i is the index type -- c is the collection type class Collection i c where getThing :: c -> i -> Thing -- A set of annotations on collection objects -- a is an annotation type constructor (a y is an annotation of type y) class Collection i c => Annotation i c a where collection :: a y -> c -- access the underlying Collection select :: a y -> i -> y -- select the annotation of an item given its index -- My custom collection data Things = Things [Thing] deriving Show -- An annotation set based on a list data MyAnnotation c y = Ann c [y] instance Collection Int Things where getThing (Things ts) i = ts !! i instance Annotation Int Things (MyAnnotation Things) where collection (Ann c _) = c select (Ann _ as) i = as !! i main = print $ select ann (2 :: Int) where ts = Things [Foo, Bar, Foo, Foo] ann = Ann ts ["a", "b", "c", "d"] -------------------- This does not work (No instance for (Annotation Int c0 (MyAnnotation Things)) arising from a use of `select'). I do not know how I can tell the type system that c0 is Things here, if it is actually possible. I'll appreciate any help or comments on how to rewrite this. Cheers, -- Olivier Iffrig

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Haskell type definitions
by Angus Comber 12 Apr '13

12 Apr '13

I am learning Haskell using Learn You a Haskell. On page 54 is an implementation of take like so: take' :: (Num i, Ord i) => i -> [a] -> [a] take' n _ | n <= 0 = [] take' _ [] = [] take' n (x:xs) = x : take' (n-1) xs I understand all the code apart from the first line. The :: part I understand as meaning this is a type definition? (Num i, Ord i) is a tuple. The first element of the tuple has to be numeric, fair enough. the second param has to be able to be ordered. The parameter is the same - both are i. This means that the types have to be the same? Why is it not (Num i, Ord j)? Isn't the 2nd tuple element referring to the list? Which could be of any type? What does => signify? i -> [a] -> [a] means first parameter is numeric? 2nd param is any type list, 3rd param is any type list. So this is saying first param numeric, 2nd param a list of any type and it returns a list of any type. Well that is understandable I suppose.

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