October 2001 - Haskell-Cafe

Re: newtype | data
by cwitty＠newtonlabs.com 05 Oct '01

05 Oct '01

Mark Carroll <mark(a)chaos.x-philes.com> writes: > Why does "newtype" exist, instead of letting people always use "data" and > still get maximum efficiency? After all, surely the implementation is an > implementation detail - a compiler could see the use of "data" with a > unary constructor and implement it as it does "newtype", instead of making > the programmer worry about how things are actually represented? > > I'm obviously missing something obvious here; I'm hoping to learn what. > (-: newtype is strict; data is not. Given > data T1 = T1 Int > newtype T2 = T2 Int > data T3 = T3 !Int you get the following results. Data> (T1 undefined) `seq` () () Data> (T2 undefined) `seq` () *** Exception: Prelude.undefined Data> (T3 undefined) `seq` () *** Exception: Prelude.undefined I can't think of a semantic difference between newtype and data with a single unary strict constructor; I suppose it would be possible to remove newtype from the language and make people declare things like T3 instead. Carl Witty

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newtype | data
by Mark Carroll 05 Oct '01

05 Oct '01

Why does "newtype" exist, instead of letting people always use "data" and still get maximum efficiency? After all, surely the implementation is an implementation detail - a compiler could see the use of "data" with a unary constructor and implement it as it does "newtype", instead of making the programmer worry about how things are actually represented? I'm obviously missing something obvious here; I'm hoping to learn what. (-: -- Mark

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broken link
by Jorge Adriano 05 Oct '01

05 Oct '01

Just thought I'd report that the PolyGP link: http://www.cs.ucl.ac.uk/staff/T.Yu/research.html at the 'Haskell in Practice' section (http://www.haskell.org/practice.html) seems to be broken. J.A.

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Re: classes and instances
by Ashley Yakeley 04 Oct '01

04 Oct '01

At 2001-10-04 08:31, Russo, Tom wrote: >Main> int 5 >ERROR - Unresolved overloading >*** Type : (Num a, ToInt a) => Int >*** Expression : int 5 > >Can someone explain what the problem is, and how to interpret the error >message I'm getting? Oh, 5 isn't necessarily an Int. It's of type "Num a => a", which is more general. You can restrict it to Int with "(5 :: Int)". -- Ashley Yakeley, Seattle WA

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classes and instances
by Russo, Tom 04 Oct '01

04 Oct '01

Hello, I'm learning haskell and I'm having some trouble with type classes. I understand them conceptually, but when I try to create an instance I run into problems. As a simple example, I create a class called ToInt: > class ToCh a where > int :: a -> Int Then I create an instance: > instance ToInt Int where: > int n = n However, when I try to use the function int :: Int -> Int at the hugs prompt, I get the following error message: Main> int 5 ERROR - Unresolved overloading *** Type : (Num a, ToInt a) => Int *** Expression : int 5 Can someone explain what the problem is, and how to interpret the error message I'm getting? thanks _t

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Re: Small confusion with monads
by Ashley Yakeley 04 Oct '01

04 Oct '01

At 2001-10-04 02:02, Dmitry Olyenyov wrote: >I have the list ["file1","file2","file3"] and I need the >function to return me the list of contents of these files. > >contents :: [String] -> [String] That cannot be safely done with that type. A "pure" function like that is not allowed to do "imperative" stuff such as reading files. -- Ashley Yakeley, Seattle WA

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Small confusion with monads
by Dmitry Olyenyov 04 Oct '01

04 Oct '01

Hello! Can anybody help me with writing function that does following: I have the list ["file1","file2","file3"] and I need the function to return me the list of contents of these files. contents :: [String] -> [String] contents (file:files) = do c <- readFile file ........ Thanks in advance for help! -- With best regards, Dmitry

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always new instance?
by Cagdas Ozgenc 04 Oct '01

04 Oct '01

Hi, Do I ALWAYS need to create a new instance if I want to modify the state of an instance? For example, if I design an index for a simple database with an recursive algebric Tree type, do I need to recreate the whole Tree if I insert or remove an element? How can I improve performance, what are common idioms in these situations? Thanks

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Re: Doing exercises from "Haskell tutorial" ...
by Marcin 'Qrczak' Kowalczyk 03 Oct '01

03 Oct '01

02 Oct 2001 15:16:27 +0300, Dmitry Astapov <adept(a)umc.com.ua> pisze: > I need to define SetsAsLists as an instance of Set by supplying > definitions for all Set methods, but definitions I wrote led me to > adding additional constraints on "union" and "memeber" methods. What constraints? The class already says that these operations require Eq on the element type: > class Set s where > empty :: s a > isEmpty :: s a -> Bool > singleton :: a -> s a > union :: Eq a => s a -> s a -> s a ^^^^ > member :: Eq a => a -> s a -> Bool ^^^^ > choice :: s a -> (a, s a) -- __("< Marcin Kowalczyk * qrczak(a)knm.org.pl http://qrczak.ids.net.pl/ \__/ ^^ SYGNATURA ZASTÊPCZA QRCZAK

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Doing exercises from "Haskell tutorial" ...
by Dmitry Astapov 02 Oct '01

02 Oct '01

It seems like an appropriate page for aske newbie questions, isnt it? I was reading through Haskell tutroial (http://www.di.uminho.pt/afp98/PAPERS/Tutorial.ps) and trying to do exercises from there. I'm stuck in the first exercise for "Classes" chapter (page 11). I need to define SetsAsLists as an instance of Set by supplying definitions for all Set methods, but definitions I wrote led me to adding additional constraints on "union" and "memeber" methods. Is it ok or it's possible to define "union" and "member" without using "==" ? Code follows: class Set s where empty :: s a isEmpty :: s a -> Bool singleton :: a -> s a union :: Eq a => s a -> s a -> s a member :: Eq a => a -> s a -> Bool choice :: s a -> (a, s a) data SetsAsLists a = SL [a] instance Set SetsAsLists where empty = SL [] isEmpty (SL []) = True isEmpty _ = False singleton x = SL [x] member x (SL ys) | isEmpty (SL ys) = False | otherwise = if x==head ys then True else member x (SL (tail ys)) union (SL []) (SL ys) = SL ys union (SL (x:xs)) (SL ys) | member x (SL ys) = union (SL xs) (SL (x:ys)) | otherwise = union (SL xs) (SL ys) -- Dmitry Astapov //ADEpt E-mail: adept(a)umc.com.ua GPG KeyID/fprint: F5D7639D/CA36 E6C4 815D 434D 0498 2B08 7867 4860 F5D7 639D

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