April 2008 - Haskell-Cafe

Re: [Haskell-cafe] [GSoC] Porting HaRe to use the GHC API
by Chaddaï Fouché 03 Apr '08

03 Apr '08

2008/4/1, Claus Reinke <claus.reinke(a)talk21.com>: > as for the project, the are actually two APIs to consider, > GHC's and HaRe's, and the main stumbling points are those > things that are not in those APIs (explicitly or at all): > > <snip> Many thanks for the valuable insights. I intend to work in close collaboration with the Kent team, indeed I'll do an internship over the summer with Simon Thompson ! Don't worry on this point. Of course I'll also try to work with the GHC developers on the subject of the GHC API. My objective is to make it work on the 6.8 API but if it appears that certain functions would be useful in the future API I'll be sure to report this to the GHC team. If I can, I would also like to be present during the discussion between the two Simon. Another notion I was interested in was to be able to reproduce a sequence of multi-module refactorings even in the absence of the initial module. It would allow to present a kind of "interface diff" for a distribution which would allow the developer of a module which depends on the distribution to update the call to this library. Of course it wouldn't be perfect and wouldn't convey the semantic modifications of the interface but it would help for all the renaming and usual modifications of external interfaces. To refactor modules that use CPP, I can see some possibilities : transforming the macro into placeholders (undefined) and inversing the transformation after the refactoring, trying every combination of the variables for the #ifdef/ifndef, forgetting those that don't compile and confronting the refactorisation of the others to see if they merge... That would need some work and would probably be very inefficient as it is But all of that should wait after we get HaRe working with the GHC API. :-) -- Chaddaï

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compile-time question
by Galchin Vasili 03 Apr '08

03 Apr '08

hello, I have added new function definitions to the namespace of an existing module. I did 1: runhaskell Setup.hs configure 2. runhaskell Setup.hs build 3) runhaskell Setup.hs install successfully so that a Linux .a(rchive) is installed. I write a small test case that references one of the new functions; however, when I build the test case, I get an "Out of scope" message indicated that at ghc(i) compile time the new function symbol cannot be found. Why? kind regards, Vasili

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Precedence (Part 2)
by PR Stanley 03 Apr '08

03 Apr '08

Hi folks > let f = sum.map (^2).filter even > f [1..4] 20 So far so good! > sum.map (^2).filter even [1..4] returns an error. How is it that the composition above as part of a function equation doesn't return any errors while on its own it requires parentheses? I can understand the need for the parentheses - the composition operator takes two arguments/operand,s each of which is a mapping which must correspond to specific domain and range requirements i.e. (b->c -> (a->b) -> (a->c) Why aren't the parentheses required in the function definition? Cheers, Paul

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Typing with co/contra-variance.
by Ben Lippmeier 03 Apr '08

03 Apr '08

Hi all, I have some quick questions for the type theory people: If I write an expression: (if .. then 23 else "Erk") In Haskell this would be an error, but perhaps I can assign it the type 'Top' (or 'Any') and then use reflection ala Data.Dynamic to inspect the type of this object at runtime and cast it back to something useful... But if I write: isEven "Erk" where isEven :: Int -> Bool isEven i = case i of 1 -> False 2 -> True ... Then this would be a genuine type error, because the case-expression demands an actual Int at runtime. Questions: 1) Could I perhaps wave my arms around and say something about the Int in the type of isEven being in a contra-variant position? Is this what is actually happening?, and can someone point me to a good paper? 2) This seems simpler than real intersection types, which I know are problematic. If I only want 'Top' and not a real object system like in Scala or Java, then can I just add this to a HM/unification style inference algorithm and expect it to work? I'm thinking the unifier only needs to know the variance of the types it's unifying to decide whether to throw an error or return Top - or will there be problems with higher order functions etc? I had a read through the subtyping chapter in "Types and Programming Languages" by Pierce - it gives the typing rules but don't mention inference or implementations. Scala seems to have this (http://www.scala-lang.org/intro/variances.html) but as I understand it, their system doesn't support full type inference without user supplied annotations... BTW: searching for 'the any type' in Google helps even less than you might expect :) Thanks, Ben.

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installing ghc-6.8.2
by michael＠schmong.org 03 Apr '08

03 Apr '08

It seems there is no ghc 6.8.2 for my version of debian. So I am compiling from source. I ran config with no problem. make generated some errors. I omit all output except the end. Let me know if Ileft something important out please. make all /usr/bin/ghc -H16m -O -w -I. -Iinclude -Rghc-timing -ignore-package Cabal -ignore-package filepath -I../libraries -fglasgow-exts -no-recomp -c System/FilePath/Posix.hs -o System/FilePath/Posix.o -ohi System/FilePath/Posix.hi cc1: error: unrecognized option `-fno-unit-at-a-time' <<ghc: 129595956 bytes, 17 GCs, 2640537/5069484 avg/max bytes residency (3 samples), 20M in use, 0.00 INIT (0.00 elapsed), 0.67 MUT (0.76 elapsed), 0.11 GC (0.14 elapsed) :ghc>> make[2]: *** [System/FilePath/Posix.o] Error 1 make[1]: *** [boot] Error 2 make: *** [stage1] Error 1 so it's this line that seems of the most interesting cc1: error: unrecognized option `-fno-unit-at-a-time' any tips on how to deal with this would be most helpful. Michael Litchard

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Type constraints for class instances
by Krzysztof Skrzętnicki 03 Apr '08

03 Apr '08

Hello everyone, I'm working on a small module for comparing things incomparable with Ord. More precisely I want to be able to compare equal infinite lists like [1..]. Obviously (1) compare [1..] [1..] = _|_ It is perfectly reasonable for Ord to behave this way. Hovever, it doesn't have to be just this way. Consider this class class YOrd a where ycmp :: a -> a -> (a,a) In a way, it tells a limited version of ordering, since there is no way to get `==` out of this. Still it can be useful when Ord fails. Consider this code: (2) sort [ [1..], [2..], [3..] ] It is ok, because compare can decide between any elements in finite time. However, this one (3) sort [ [1..], [1..] ] will fail because of (1). Compare is simply unable to tell that two infinite list are equivalent. I solved this by producing partial results while comparing lists. If we compare lists (1:xs) (1:ys) we may not be able to tell xs < ys, but we do can tell that 1 will be the first element of both of smaller and greater one. You can see this idea in the code below. --- cut here --- {-# OPTIONS_GHC -O2 #-} module Data.YOrd where -- Well defined where Eq means equality, not only equivalence class YOrd a where ycmp :: a -> a -> (a,a) instance (YOrd a) => YOrd [a] where ycmp [] [] = ([],[]) ycmp xs [] = ([],xs) ycmp [] xs = ([],xs) ycmp xs'@(x:xs) ys'@(y:ys) = let (sm,gt) = x `ycmp` y in let (smS,gtS) = xs `ycmp` ys in (sm:smS, gt:gtS) ycmpWrap x y = case x `compare` y of LT -> (x,y) GT -> (y,x) EQ -> (x,y) -- biased - but we have to make our minds! -- ugly, see the problem below instance YOrd Int where ycmp = ycmpWrap instance YOrd Char where ycmp = ycmpWrap instance YOrd Integer where ycmp = ycmpWrap -- ysort : sort of mergesort ysort :: (YOrd a) => [a] -> [a] ysort = head . mergeAll . wrap wrap :: [a] -> [[a]] wrap xs = map (:[]) xs mergeAll :: (YOrd a) => [[a]] -> [[a]] mergeAll [] = [] mergeAll [x] = [x] mergeAll (a:b:rest) = mergeAll ((merge a b) : (mergeAll rest)) merge :: (YOrd a) => [a] -> [a] -> [a] merge [] [] = [] merge xs [] = xs merge [] xs = xs merge (x:xs) (y:ys) = let (sm,gt) = x `ycmp` y in sm : (merge [gt] $ merge xs ys) --- cut here --- I'd like to write the following code: instance (Ord a) => YOrd a where ycmp x y = case x `compare` y of LT -> (x,y) GT -> (y,x) EQ -> (x,y) But i get an error "Undecidable instances" for any type [a]. Does anyone know the way to solve this? Best regards Christopher Skrzętnicki

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ANN: hdbc-odbc 1.1.4.1
by John Goerzen 02 Apr '08

02 Apr '08

Hi folks, HDBC-ODBC 1.1.4.1 has been uploaded to http://software.complete.org/hdbc-odbc and to Hackage. Bryn Keller reported a build problem on Windows with GHC 6.8.x, which has been fixed. -- John

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FW: Haskell
by Simon Peyton-Jones 02 Apr '08

02 Apr '08

Dear Haskell Cafe members Here's an open-ended question about Haskell vs Scheme. Don't forget to cc Douglas in your replies; he may not be on this list (yet)! Simon -----Original Message----- From: D. Gregor [mailto:kerrangster@gmail.com] Sent: 30 March 2008 07:58 To: Simon Peyton-Jones Subject: Haskell Hello, In your most humble opinion, what's the difference between Haskell and Scheme? What does Haskell achieve that Scheme does not? Is the choice less to do with the language, and more to do with the compiler? Haskell is a pure functional programming language; whereas Scheme is a functional language, does the word "pure" set Haskell that much apart from Scheme? I enjoy Haskell. I enjoy reading your papers on parallelism using Haskell. How can one answer the question--why choose Haskell over Scheme? Regards, Douglas

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Template Haskell -- when are things evaluated?
by Emil Axelsson 02 Apr '08

02 Apr '08

Hello all! Up until yesterday I thought I understood the basics of Template Haskell, but now I'm a little confused. Consider the following code module A where a1 = [| (2::Int) + 2 |] a2 = let x = (2::Int) + 2 in [| x |] a3 = [| y |] where y = (2::Int) + 2 z = (2::Int) + 2 a4 = [| z |] module B where import A a1S = $a1 a2S = $a2 a3S = $a3 a4S = $a4 I'd have thought that in all four cases the addition was evaluated at compile-time, but compiling with -ddump-splices reveals that this is only the case for a2 and a3. Is there a general reliable rule for when things are evaluated? Thanks, / Emil

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Re: [Haskell-cafe] Function Precedence
by PR Stanley 02 Apr '08

02 Apr '08

>Think about this: > >map (+1) [1..10] > >What should it do? > take (+1) and return a function which takes a list as its > argument and finally return a list. >How about: > >f 1 2 3 > >Should that be f (1 (2 3)), or ((f 1) 2) 3? The latter, of course, but that's not really what I'm driving at. I'm asking why we can't have a function treated differently with regard to the precedence and associativity rules. f 1 2 is indeed ((f 1) 2). Why not f 1 g 2 == ((f 1) (g 2))? Cheers, Paul

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