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composeList
by Emre Tezel 13 May '02

13 May '02

Hi all, I recently bought Simon Thompson's Haskell book. I have been doing the exercises while I read on. There are couple questions that I can not solve. Any help would be greatly appreciated. I got stuck on this question in Chapter 10. Define a function composeList which composes a list of functions into a single function. What is the type of composeList? I naively tried the following but the hugs compiler complained about the inferred type not being generic enough. composeList :: [(a -> b)] -> (c -> b) composeList [] = id composeList (x:xs) = x . (composeList xs) or composeList :: [(a -> b)] -> (c -> b) composeList (x:xs) = x . (composeList xs) ERROR "Chapter10.hs":6 - Inferred type is not general enough *** Expression : composeList *** Expected type : [a -> b] -> c -> b *** Inferred type : [a -> a] -> b -> a Anybody can solve this? Emre

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Re: waiting for your reply
by Bob Johnson 12 May '02

12 May '02

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preprocessing printf/regex strings (like ocaml)
by Oliver George 12 May '02

12 May '02

Hi Haskellers, I'm interested to know why a string translating preprocessor doesn't exist for haskell. It seems to me that it would alleviate printf and regex like problems in an convenient, elegant and type-safe manner. An example of this I came across recently was the ocaml printf statement: # Printf.printf "roar";; roar- : unit = () # Printf.printf "numbers %d %d" 11 23;; numbers 11 23- : unit = () # Printf.printf "a number %d" "word";; This expression has type string but is here used with type int You can see logically how this might work in haskell, compile time evaluation would translate the string into the appropriate native expressions before the statements type is evaluated. I see a similar possibility for regular expression matching, wouldn't it be nice in a haskell script to write: perl like things... msg' = replace "s/love/lust/" msg or, nice regex stuff... main = case match "^(\d+)" of Nothing -> 0 Just (i) -> i the python string notation (str % tuple) would fit really well too... putStrLn "hello %s, you got %d right" % ("oliver", 5) Am I the only one who sees this as being a really valuable extension to haskell? or does it exist and i've just never noticed? cheers, Oliver.

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Wishing you a wonderful day
by Bob Johnson 12 May '02

12 May '02

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examples of functional objects
by Pixel 11 May '02

11 May '02

It seems to be hard to have something both abstract and concrete. http://merd.net/pixel/language-study/various/non-mutable-objects-and-subtyp… The first version (file haskell.listing) is quite simple, but the last one (file haskell3.listing) gets pretty hairy: ---------------------------------------- class Show a => Point a where translate :: a -> Double -> a my_length :: a -> Double add :: a -> a -> a pack :: a -> Packed_point unpack :: a -> Points pack = Packed_point data Packed_point = forall a. Point a => Packed_point a data Points = TPoint2d Point2d | TPoint3d Point3d data Point2d = Point2d(Double, Double) deriving Show data Point3d = Point3d(Double, Double, Double) deriving Show ---------------------------------------- is there a simpler solution? thanks! PS: please, don't give me the simple non-OO version, it's already there (haskell3bis.listing), and I give an explaination why i think it's not enough: non-OO Haskell and OCaml versions are the simplest, but do not allow precise enough typing: no distinction between Point2d and Point3d is done at compile time. So the only way to have a function working only on Point3d's is to raise an error when that function is called with a Point2d (see get_z) (of course, compared to dynamically typed languages versions, it's already much better)

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ObjectIO install
by Andre W B Furtado 10 May '02

10 May '02

The INSTALL file that comes with the objectIO files says that one must use the following commands to register the package: $ ghc-pkg -r objectio $ ghc-pkg -a objectio -f pckg.cfg The second command is not recognized as a valid one, and the ghc-pkg help appears showing the correct syntax. I needed to use $ ghc-pkg -a < pckg.cfg in order to get the package correctly registered. Am I missing something or is the INSTALL file wrong? -- Andre

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by sales137＠163.com 09 May '02

09 May '02

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RE: Using "type"
by Simon Peyton-Jones 09 May '02

09 May '02

GHC currently takes the following (non-Haskell 98) attitude to type synonyms: first expand type synonyms only then check for partial applications of type synonyms then perform type checking So in his case type for f expands to: f :: m () -> m () test :: Int -> Int You can see how the type for test comes out: Thing (Const Int) = Const Int () = Int Now GHC tries to unify m () with Int and fails, as it should. The error message is phrased in terms of the original un-expanded types, which is usually a good thing but on this occasion gives a rather odd looking error message. GHC shouldn't really do this lazy checking unless you say -fglasgow-exts Simon | -----Original Message----- | From: Samuel E. Moelius III [mailto:usmoeliu@mcs.drexel.edu] | Sent: 04 May 2002 20:45 | To: haskell-cafe(a)haskell.org | Subject: Using "type" | | | I apologize if this question has been asked before... | | It seems to me the following code should be legal: | | type Thing m = m () | | type Const a b = a | | f :: Thing m -> Thing m | f x = x | | test :: Thing (Const Int) -> Thing (Const Int) | test = f | | However, GHC gives the following error: | | Couldn't match `Thing m' against `Thing (Const Int)' | Expected type: Thing m | Inferred type: Thing (Const Int) | Expected type: Thing (Const Int) -> Thing (Const Int) | Inferred type: Thing m -> Thing m | | But if you change the definition of Const | | newtype Const a b = MakeConst a | | the code compiles fine. | | Is this the desired behavior? | | Is it the case that when you have quantification of a type | variable of | kind * -> * (in this case, m), that variable may only become | bound to a | type constructor? | | Sam Moelius | | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe(a)haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe |

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RE: Deforestration harmful?
by Simon Marlow 08 May '02

08 May '02

> I will be a bit more precise. Let f and g have type: > > g :: Seed -> [Tree] > f :: Tree -> [Expr] > > where Tree1 is a tip-valued binary tree, and Tree2 is also a > binary tree but the forks are arithmetic operators: > > data Tree = Tip Int | Bin Tree1 Tree1 > data Expr = Val Int | Add Tree2 Tree2 | Sub Tree2 Tree2 | Mul .... These don't make sense. Perhaps for Tree1 you mean Tree, and Tree2 you mean Expr? > The function g is defined like: > > g seed = [ z | (s1, s2) <- split seed, > x <- g s1, y <- g s2, z <- Bin x y ] There's a type error in that definition: 'Bin x y' is not a list. > The function f converts a Tree to many Exprs by replacing > Bin with one of Add, Sub, Mul or Div: > > f (Tip i) = [Val i] > f (Bin x y) = [op x' y' | x' <- f x, y' <- f y, op <- join x y] > > The program before fusion looks like: > > concat . map f . g > > while the fused program simply: > > fg seed = [ z | (s1, s2) <- split seed, > x <- fg s1, y <- fg s2, z <- join x y ] > join x y = [op x y | op <- [Add, Sub, ... ]] > > And the surprise was the fused program ran slower than the > naive one before fusion. > > The time profiling showed that, in one run, the naive > program itself took 2.00 sec to finish, with 9% extra GC time. The > fused program itself, on the other hand, took only 1.84 sec but > spent an extra 67% of time on GC (the heap size is the default > 64M). > > Considering their heap profiles, both of them look like several > steep, slanted triangles standing next to each other. The highest peak > in the heap profile of the naive program reaches only 50K of memory, > while that for the fused program reaches as much as 1500K. In > the former case, the heap is mostly occupied by chunks created > by g. In the latter case, the heap is mostly occupied by join. > > > Furthermore, deforestation itself does not reduce runtime very much. > > Indeed. Looks like it helps to speed up the mutator a bit but > end up spending more time on GC, due to increased heap demand. I think this is a misleading generalisation - in most cases deforestation doesn't affect the heap residency behaviour of the program. As Olaf said, deforestation tends to result in small constant factor speedups, but can be particularly beneficial if it enables further optimisations to take place. In your case it looks like the transformation you made by hand had the side-effect of introducing a space leak, but it's quite hard to tell from the code you've shown us. (space leaks are particularly hard to identify by eye, using the profiling tools is the only way to really understand what's happening). > Monday, May 6, 2002, 12:02 PM, Simon Marlow wrote： > > It is entirely plausible that as you increase the heap > size the mutator > > time increases: this is because you get fewer cache hits > with a larger > > heap. GHC starts off with a small heap size so as to try > to make the > > most of the cache. However, on most examples we've seen > the benefits of > > reducing garbage collections by increasing the heap size tend to > > outweigh the benefits from staying in the cache. > > Did you try heap profiling to discover what structure was > filling up > > the heap? > > Thanks for the explanation. :) > > More info. about the heap was given above. By the way, could there be > situations where increasing the heap size also increases the GC time? Yes, again because of increased cache misses. Cheers, Simon

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Re: Proper scaling of randoms
by Dylan Thurston 07 May '02

07 May '02

(Redirected to haskell-cafe.) On Mon, May 06, 2002 at 04:49:55PM -0700, oleg(a)pobox.com wrote: > Problem: given an integer n within [0..maxn], design a scaling > function sc(n) that returns an integer within [s..t], t>=s. > The function sc(n) must be 'monotonic': > 0<=a < b ==> sc(a) <= sc(b) > and it must map the ends of the interval: > sc(0) -> s and sc(maxn) -> t. Just an aside (which Oleg surely knows): for actual random number generation, you often don't care about the monotonicity, and only care about uniform generation. In this case, there is a very simple algorithm: work modulo (s-t+1). scm(n) = (n `rem` (s-t+1)) + s Warning: some, broken, random number generators do not behave well when used like this. Also, although this is as uniform as possible, there is a systematic skew towards the lower end of the range [s..t]. --Dylan Thurston

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