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The difference between ($) and application
by oleg＠pobox.com 14 Dec '04

14 Dec '04

The operator ($) is often considered an application operator of a lower precedence. Modulo precedence, there seem to be no difference between ($) and `the white space', and so one can quickly get used to treat these operators as being semantically the same. However, they are not the same in all circumstances. I'd like to observe an important case where replacing the application with ($) in a fully-parenthesized expression can lead to a type error. > {-# OPTIONS -fglasgow-exts #-} > > module Foo where > > data WR = WR (Int -> Int) > data W = W (forall a. a->a) > > t1 = WR id > t2 = W id We can also write > t1' = WR $ id However, if we try > t2' = W $ id we get an error: /tmp/t1.hs:13: Inferred type is less polymorphic than expected Quantified type variable `a' escapes Expected type: (a -> a) -> b Inferred type: (forall a1. a1 -> a1) -> W In the first argument of `($)', namely `W' In the definition of `t2'': t2' = W $ id Incidentally, Hugs -98 gives a quite bizarre error message ERROR "/tmp/t1.hs":13 - Use of W requires at least 1 argument It didn't complain about "WR $ id"... The reasons for that behavior are obvious: the compiler cannot generalize to higher-ranked types when instantiating the type of ($). It makes a difference that the application is a built-in construction, whereas ($) is just a regular function.

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special term describing f :: (Monad m) => (a -> m b) ?
by Henning Sato von Rosen 14 Dec '04

14 Dec '04

What is a function of the followning type called: f :: (Monad m) => (a -> m b) Is there a special term describing such a function (a function into a monad)? For f in a >>= f is en example. Need it for an article/report. Regards/Henning

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local loops
by Per Larsson 14 Dec '04

14 Dec '04

I often use local loops in monadic code, e.g. main = do ... let loop = do ... if cond then loop else return () loop It seems that I can encode this idiom slightly more concise with the 'fix' operator (from Control.Monad.Fix), i.e. main = do ... fix (\loop -> do ... if ...) But is it really semantically equivalent? Is it as efficient? Per

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Flattening tail recursion?
by GoldPython 13 Dec '04

13 Dec '04

I'm missing something, a functional idiom, whatever. I know there's "length" to count the elements in a list and it works fine, but the function below stack overflows on a large list. countLines [] = 0 countLines (_:ls) = 1 + countLines ls I would have thought that this was tail recursive and would be flattened into iteration by the compiler. Can anyone explain why? Is it because the call is embedded in an expression? And additionally, how can I predict when the flattening will occur? Thanks, Will

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FiniteMapFiniteMap
by Dominic Steinitz 13 Dec '04

13 Dec '04

See also http://www.mail-archive.com/haskell-cafe@haskell.org/msg05809.html Dominic.

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FiniteMapFiniteMap
by S. Alexander Jacobson 13 Dec '04

13 Dec '04

I would like to read a large finitemap off of a disk faster than the time it takes to read the entire list of pairs. My solution is to save it as a bunch of smaller lists of pairs covering various key intervals (or recency intervals). I then can readfile all of these lists back into a bunch of finitemaps. Because readfile is lazy my program can start very quickly without waiting to read the contents of all of these files. To make this solution work, I need some datastructure that looks like a finitemap but operates on a bunch of them organized either in parallel (key interval model) or series (time interval model). Is there some pre-existing data structure that does this or do I need to roll my own? Or is there a better way to (de-)serialize FiniteMaps? -Alex- ______________________________________________________________ S. Alexander Jacobson tel:917-770-6565 http://alexjacobson.com

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AbstractDataType question
by Jason Bailey 13 Dec '04

13 Dec '04

So in the wiki on AbstractDataType ( http://haskell.org/hawiki/AbstractDataType ) it gives the following example: data Tree a = Nil | Node { left :: Tree a, value :: a, right :: Tree a } The part where I am confused is with the use of '{' and '}' my reference book is apparently out of date as it states that Haskell doesn't even use those characters :) Could someone point me to some documentation on whats being created here and how its used? thank you Jason

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RE: [Haskell-cafe] Flattening tail recursion?
by Simon Peyton-Jones 13 Dec '04

13 Dec '04

I have not been following the details, but yes, the strictness analyser only runs when you say -O, and without strictness analysis you can get different space behaviour. Sometimes more, sometimes less. In this case, strictness analysis helps. Simon | -----Original Message----- | From: haskell-cafe-bounces(a)haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of Jules | Bean | Sent: 10 December 2004 21:38 | To: GoldPython | Cc: Henning Thielemann; haskell-cafe | Subject: Re: [Haskell-cafe] Flattening tail recursion? | | | On 10 Dec 2004, at 20:33, GoldPython wrote: | | > Just compiled this with -O and it ran with no stack overflow. | > Evidently, no `seq` needed for this one. Using ghc 6.2.2. | > | > countLines l = countLines' 0 l | > countLines' n [] = n | > countLines' n (_:ls) = countLines' (n + 1) ls | > | | That's presumably the answer. GHC's strictness analyser *can* cope with | this situation but only under -O... whereas most of us where testing | under ghci... | | Confirmation from one of the Simons? | | Jules | | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe(a)haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe

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ghc install
by Radu Grigore 11 Dec '04

11 Dec '04

Hi, I try to install ghc on 6.2 on a gentoo linux distribution (from sources). It fails with --- /usr/bin/ar: creating libHSbase.a xargs: /usr/bin/ar: terminated by signal 15 --- It looks like it's a lack of resources. The computer is a Celeron 1.3GHz with 160MB RAM & 400MB swap. Isn't that enough to compile ghc? -- regards, radu http://rgrig.idilis.ro/

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Class Synonyms
by Jorge Adriano Aires 11 Dec '04

11 Dec '04

Hello! I got a multi-parameter type class: > class Foo a b | a -> b where > foo_method1 :: ... > foo_method2 :: ... > ... And some particular cases are important on their own, like the one where 'a' and 'b' are the same, I call elements with this property, Bar. So I defined: class Foo a a => Bar a where This is nice, now I can replace (F a a) for (Bar a) in the context of many functions. Less typing and it's more readable. Still, I have to define instances as: > instance Foo A A where > foo_method1 = bla1 > foo_method2 = bla2 > > instance Bar A where I'd like to achieve something better than this. I was looking for something like "class synonyms". So if I declared an instance Foo A A, I automagicaly had Bar A, because they'd be the same. Or even better, I could declare > instance Bar A where > foo_method1 = bla1 > foo_method2 = bla2 Is anything like this possible at all? Thanks, J.A.

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