June 2003 - Haskell - Haskell.org

[tim@epicgames.com: Fw: Typesafe MRef with a regular monad]
by Andrew J Bromage 05 Jun '03

05 Jun '03

G'day all. On Wed, Jun 04, 2003 at 07:31:03PM -0500, Tim Sweeney wrote: > Conjecture: It's impossible to implement RefMonad directly in Haskell > without making use of built-in ST or IO functionality and without unsafe or > potentially diverging code (such as unsafeCoerce). > Any takers? WARNING: This is not rigorous. Take a prototypical simple state monad: type MyRefMon a = SomeState -> (a. SomeState) type MyRef a = Int -- Could be anything other than Int, but -- the important point is that "a" is not -- mentioned on the RHS anywhere. where SomeState is abstract but concrete. Note that it has no type parameters. Let's also assume the existence of this function: myDeref :: MyRef a -> MyRefMon a Again, we know nothing about its implementation (yet). Without loss of generality, and because a has no type constraints, myDeref is equivalent to a function of this type: myDeref' :: Int -> SomeState -> (a, SomeState) (This step I believe to be far too handwavy. I have no justification for the ability to drop the "a" parameter here except that Haskell semantics state that if a is not constrained, you can't do anything with it.) Intuitively, there is nothing in the above type signature which gives you any way to construct something of type a, therefore it must be bottom. The parametricity theorem (i.e. "free theorem") can probaly be used to show this formally. The only ways I can think of to get over this within Haskell is to: - change the definition of MyRef so that it contains enough information to extract something of type a from SomeState, or - constrain the type (e.g. using Typeable). Cheers, Andrew Bromage

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RE: Haddock module description
by Graham Klyne 04 Jun '03

04 Jun '03

At 14:58 04/06/03 +0100, Simon Marlow wrote: >Haddock understands the style of module header that we're using for the >hierarchical libraries. The module header is described in this document >(see the section Reference Libraries->Coding Style->Module Header): > >http://www.haskell.org/hierarchical-modules/libraries/libraries.html Great... just what I wanted. Thanks. #g ------------------- Graham Klyne <GK(a)NineByNine.org> PGP: 0FAA 69FF C083 000B A2E9 A131 01B9 1C7A DBCA CB5E

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forall quantifier
by ketil＠ii.uib.no 04 Jun '03

04 Jun '03

Hi, This is one of those topics everybody else seems to be familiar with, but which I don't quite understand, and can't seem to find any good information about. I have a function declared as: anova2 :: (Fractional c, Ord b) => [a->b] -> (a->c) -> [a] -> [Anova1 c] where the first parameter is a list of classifiers. I could simplify it, I guess, to something like classify :: Eq b => [a->b] -> [a] -> [[[a]]] classify cs xs = ... where for each classifying function in cs, I would get the xs partitioned accordingly. E.g. classify [fst,snd] [(1,0), (1,2), (2,0)] would yield [ [(1,0), (1,2)], [(2,0)] -- classified by `fst` , [(1,0), (2,0)], [(1,2)]] -- classified by `snd` Now, obviously, the problem is that fst and snd, being passed in a list, needs to be of the same type; this complicates classifying a list of type [(Int,Bool)], for instance¹. I have a vague notion this is solvable using quantifiers (since I ever only use Eq operations on the type), but I'm not sure exactly how, I can't seem to find a good tutorial, and my Monte-Carlo programming approach doesn't seem to be leading anywhere :-) Can somebody suggest a solution, or a place to look? -kzm ¹ I guess I can convert Bool to Int (True->1, False->0), but it's not very appealing, IMHO. -- If I haven't seen further, it is by standing in the footprints of giants

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RE: forall quantifier
by Markus.Schnell＠infineon.com 04 Jun '03

04 Jun '03

Yes, this would work, thanks. But let me extent my question: what if all the types would be in a class FakeClass which has function specialID :: a -> ID and I would like to do > foo $ map specialID [a,b,c,d] ?

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RE: Haddock module description
by Simon Marlow 04 Jun '03

04 Jun '03

> The Haddock document [1] section 3.3 gives a very simple > specification for > the module description, but the accompanying examples [2][3] > suggest that > certain common layout in the module description is picked up > and formatted > specially. Is there a description of this? Haddock understands the style of module header that we're using for the hierarchical libraries. The module header is described in this document (see the section Reference Libraries->Coding Style->Module Header): http://www.haskell.org/hierarchical-modules/libraries/libraries.html Cheers, Simon

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Re: labelled fields
by George Russell 04 Jun '03

04 Jun '03

Steffen wrote > I need something like > > data Type = TCon String {lmtc::Maybe String} ... > > but this does not seem to be possible. Instead > I have to waste identifiers: > > data Type = TCon {senseless::String,lmtc::Maybe String} ... > > Why? Are there any formal reasons against the > above declaration? Clearly it's arguable, since SML allows general record types. My SML syntax is rusty but I think you could write: > datatype ty = ty string {lmtc : string option} or indeed use {lmtc : string option} just like a type. I don't think haskell(a)haskell.org is suitable for lengthy discussions of the merits of SML versus Haskell or vice-versa, so I will state my view and leave it at that: (1) the Haskell approach costs you slightly more typing. (2) on the other hand the error messages are slightly better, since if you type TCon {sensless = "foo"} the compiler should spot immediately that "sensless" has no place in a TCon, while in ML the compiler will have to say something like "Expecting a {senseless :: String}, found a {sensless :: String}", which leaves you to work out what it is about the context that requires a "{senseless :: String}" (3) the Haskell approach allows several constructors to have the same field name, for example > data Ty1 = > TCon1 {lmtc :: Maybe String} > | TCon2 {lmtc :: Maybe String, extraField :: String} and you then have a function lmtc :: Ty1 -> Maybe String which works for all constructors having the lmtc field. (4) Haskell will fill in undefined fields, with values that give error messages if you try to look at them. Which you prefer is entirely up to you. None of the above points are terribly important. I personally prefer the Haskell approach, but not by much.

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RE: forall quantifier
by Markus.Schnell＠infineon.com 04 Jun '03

04 Jun '03

> Now, obviously, the problem is that fst and snd, being passed in a > list, needs to be of the same type; this complicates classifying a > list of type [(Int,Bool)], for instance¹. I have a similar problem. Say I have a record: > data Rec = Rec { a :: Int, b :: Type1, c :: Type2, d :: Type3, ... } > rec = Rec { a = 1, b = ... } Now I want to print part of the record. What I would like to do is the following > putStrLn $ concatMap show [a,c,d] !!! bang !!! So what I actually do is > putStrLn $ concat [show a, show c, show d] Works, but a little bit clumsy, especially with long lists. Is there a nice solution? Markus

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Haddock module description
by Graham Klyne 04 Jun '03

04 Jun '03

The Haddock document [1] section 3.3 gives a very simple specification for the module description, but the accompanying examples [2][3] suggest that certain common layout in the module description is picked up and formatted specially. Is there a description of this? (I'm starting to put simple Haddock markup into my code, but I'm not running Haddock yet so I'm not yet ready to "suck it and see".) #g -- [1] http://www.haskell.org/haddock/docs/x355.html [2] http://www.haskell.org/haddock/Test.hs [3] http://www.haskell.org/haddock/Test/Test.html ------------------- Graham Klyne <GK(a)NineByNine.org> PGP: 0FAA 69FF C083 000B A2E9 A131 01B9 1C7A DBCA CB5E

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Implementing RefMonads in Haskell without ST,IO
by Tim Sweeney 04 Jun '03

04 Jun '03

Given the common definition of RefMonad: class Monad m => RefMonad m r | m -> r where newRef :: a -> m (r a) readRef :: r a -> m a writeRef :: r a -> a -> m () Is it possible to actually implement a working instance of RefMonad in Haskell, without making use of a built-in monad like IO or ST? If so, I'd love any tips -- I've been making good use of monads for a while, but can't figure this one out. The Java programmer in me wants to implement RefMonad by passing around a function from integers (think of them as pointers or heap indices) to "objects", and in readRef, "cast" the "object" to the appropriate type "t". If it's not possible to implement a typesafe RefMonad instance directly in Haskell, without making use of built-in imperative features like IO, then doesn't this refute the claims that monads implement imperative features functionally? -Tim

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labelled fields
by Steffen Mazanek 03 Jun '03

03 Jun '03

Hello, I need something like data Type = TCon String {lmtc::Maybe String} ... but this does not seem to be possible. Instead I have to waste identifiers: data Type = TCon {senseless::String,lmtc::Maybe String} ... Why? Are there any formal reasons against the above declaration? Or do I miss something? Thanks, Steffen -- Steffen Mazanek - www.steffen-mazanek.de - GPG: 791F DCB3 Haskell, that's where I just curry until fail, unwords any error, drop all undefined, maybe break, otherwise in sequence span isControl and take max $, id: (d:[])

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