April 2003 - Hugs-Users

undefined type prob
by James Grist 01 May '03

01 May '03

I've got a prob with the Function toHeap. It return an argument with an undefined type. Well, I think that's got something to do with why I can`t get it to work. OK, for it's input, it takes a list of some datatype, so I try at the hugs console.... toHeap "hello" ERROR - Unresolved overloading *** Type : Heap a => a Char *** Expression : toHeap "hello" How do I get round this prob? The fact that this code compiles fine, makes me think that it's me not using it right, rather than the code not being right *** module Heap where class Heap h where empty :: Ord a => h a isEmpty :: Ord a => h a -> Bool insert :: Ord a => a -> h a -> h a merge :: Ord a => h a -> h a -> h a findMin :: Ord a => h a -> Maybe a deleteMin :: Ord a => h a -> h a toHeap :: (Ord a, Heap h) => [a] -> h a toHeap xs = foldr insert empty xs data Way = L | R deriving (Eq, Show) data Tree a = Nil | Node Way a (Tree a) (Tree a) deriving Show isNil :: Tree a -> Bool isNil Nil = True isNil _ = False isNode :: Tree a -> Bool isNode = not . isNil leftSub :: Tree a -> Tree a leftSub Nil = error "leftSub" leftSub (Node _ _ lt _) = lt rightSub :: Tree a -> Tree a rightSub Nil = error "rightSub" rightSub (Node _ _ _ rt) = rt root :: Tree a -> a root Nil = error "root" root (Node _ v _ _) = v insTree :: Ord a => a -> Tree a -> Tree a insTree val Nil = Node L val Nil Nil -- L is an arbitrary choice insTree val (Node way v lt rt) | v==val = Node way v lt rt -- no change, value in tree | val < v = if (way==L) then Node R val (insTree v lt) rt else Node L val lt (insTree v rt) | v < val = if (way==L) then Node R v (insTree val lt) rt else Node L v lt (insTree val rt) minTree :: Ord a => Tree a -> Maybe a minTree t | isNil t = Nothing | otherwise = Just(root t) deleteM :: Ord a => Tree a -> Tree a deleteM Nil = error "deleteM" deleteM (Node _ _ lt rt) = join lt rt join :: Ord a => Tree a -> Tree a -> Tree a join t Nil = t join Nil t = t join lt@(Node way1 v1 lt1 rt1) rt@(Node way2 v2 lt2 rt2) | v1 <= v2 = Node L v1 lt1 (join rt1 rt) | v2 < v1 = Node R v2 (join lt lt2) rt2 --instance (Show a) => Show (Tree a) where -- show Tree Nil = "as" instance Heap Tree where empty = Nil isEmpty = isNil insert = insTree merge = join findMin = minTree deleteMin = deleteM __________________________________________________ Yahoo! Plus For a better Internet experience http://www.yahoo.co.uk/btoffer

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show tree
by James Grist 30 Apr '03

30 Apr '03

ok, I've got a prob with this 'ere piece of code there is no Show instance for Tree, so I anticipated the result of typing... Nil at the hugs console, which was... ERROR - Cannot find "show" function for: *** Expression : Nil *** Of type : Tree a OK so far, next I tried to write a show instance for Tree ( near bottom, commented out here ), and I was suprised to get this error at the hugs console when I tried to compile it..... ERROR C:\My Documents\com2020\Heap.hs:16 - Overlapping instances for class "Show" *** This instance : Show (Tree a) *** Overlaps with : Show (Tree a) *** Common instance : Show (Tree a) I'm guessing that the compiler has decided that I've tried to declare a show instance for Tree twice. But if this is the case, why does typing "Nil" cause the error it does? Someone please explain. Thanks *** module Heap where class Heap h where empty :: Ord a => h a isEmpty :: Ord a => h a -> Bool insert :: Ord a => a -> h a -> h a merge :: Ord a => h a -> h a -> h a findMin :: Ord a => h a -> Maybe a deleteMin :: Ord a => h a -> h a toHeap :: (Ord a, Heap h) => [a] -> h a toHeap xs = foldr insert empty xs data Way = L | R deriving (Eq, Show) data Tree a = Nil | Node Way a (Tree a) (Tree a) deriving Show isNil :: Tree a -> Bool isNil Nil = True isNil _ = False isNode :: Tree a -> Bool isNode = not . isNil leftSub :: Tree a -> Tree a leftSub Nil = error "leftSub" leftSub (Node _ _ lt _) = lt rightSub :: Tree a -> Tree a rightSub Nil = error "rightSub" rightSub (Node _ _ _ rt) = rt root :: Tree a -> a root Nil = error "root" root (Node _ v _ _) = v insTree :: Ord a => a -> Tree a -> Tree a insTree val Nil = Node L val Nil Nil -- L is an arbitrary choice insTree val (Node way v lt rt) | v==val = Node way v lt rt -- no change, value in tree | val < v = if (way==L) then Node R val (insTree v lt) rt else Node L val lt (insTree v rt) | v < val = if (way==L) then Node R v (insTree val lt) rt else Node L v lt (insTree val rt) minTree :: Ord a => Tree a -> Maybe a minTree t | isNil t = Nothing | otherwise = Just(root t) deleteM :: Ord a => Tree a -> Tree a deleteM Nil = error "deleteM" deleteM (Node _ _ lt rt) = join lt rt join :: Ord a => Tree a -> Tree a -> Tree a join t Nil = t join Nil t = t join lt@(Node way1 v1 lt1 rt1) rt@(Node way2 v2 lt2 rt2) | v1 <= v2 = Node L v1 lt1 (join rt1 rt) | v2 < v1 = Node R v2 (join lt lt2) rt2 --instance (Show a) => Show (Tree a) where -- show a = "as" instance Heap Tree where empty = Nil isEmpty = isNil insert = insTree merge = join findMin = minTree deleteMin = deleteM *** __________________________________________________ Yahoo! Plus For a better Internet experience http://www.yahoo.co.uk/btoffer

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Hugs and cygwin
by sigurd＠12move.de 29 Apr '03

29 Apr '03

Hi, I wanted to take a look at Haskell and thougt Hugs would be a good tool to start with. I have a windows as OS so I first downloaded the windows binaries. But they didn't play well with my editor (XEmacs). In XEmacs I have a command to send the content of a buffer to a hugs process; that command saves the buffer and calls in a running hugs process the command `:load' with the file name as parameter. But the windows version didn't understand the pathnames from my XEmacs (it is build under cygwin and for that uses unix path notation). So what to do? I could frob all functions in hugs mode that deal with sending filenames to send a pathname understood by a windows version of hugs. I didn't like that idea and tried to build from the source. It worked OOTB (and now I had readline in hugs) but there is one minor problem. If I use the `:edit' command with absolute pathnames the gets mangled in a weird way. Suppose I work on c:\ . If I start a session all looks fine ,---- | $ hugs -E"winclient %s" | __ __ __ __ ____ ___ _________________________________________ | || || || || || || ||__ Hugs 98: Based on the Haskell 98 standard | ||___|| ||__|| ||__|| __|| Copyright (c) 1994-2002 | ||---|| ___|| World Wide Web: http://haskell.org/hugs | || || Report bugs to: hugs-bugs(a)haskell.org | || || Version: November 2002 _________________________________________ | | Haskell 98 mode: Restart with command line option -98 to enable extensions | | Reading file "/usr/local/lib/hugs/lib/Prelude.hs": | | Hugs session for: | /usr/local/lib/hugs/lib/Prelude.hs `---- If I say now `edit' the editor is called with /c/usr/local/lib/hugs/lib/Prelude.hs as filename. if I work on another drive lets say e:\ the filename gets /e/usr/local/lib/hugs/lib/Prelude.hs But if I use a relative pathname; eg `:load ./foo.hs' and say after that `edit' all is fine. What can I do to change that behaviour? bye KP -- 'Twas brillig, and the slithy toves Did gyre and gimble in the wabe; All mimsy were the borogoves, And the mome raths outgrabe. "Lewis Carroll" "Jabberwocky"

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Arkadasa yardimci olalim
by Dr. Ahmet Yalcin 27 Apr '03

27 Apr '03

Arkadaslar, lütfen dikkate alin, yardimci olalim: Merhaba, cok zor durumda ve baska caresi kalmamis biri olarak hosgörünüze ve insan sevginize siginarak sizden bana cok kisada olsa kulak vermenizi rica edecegim. Ben, 34 yasinda evli ve 4 yasinda olan bir cocuk babasiyim. Aile hayatimizin sorunsuz ve mutlu sekilde devam ederken biricik oglumuz aniden rahatsizlandi. Uzun süren arastirmalar sonucu oglumuz Ali Can'in kalbinde hayati bir fonksiyon bozuklugunun bulundugu doktorlar tarafindan tespit edildi. Oglumuz cok acil sekilde ameliyat olmazsa maalesef hayata gözlerini yumacak. Ancak ne varki ameliyat icin gerekli olan parayi ödememiz bizim icin mümkün gözükmüyor. Bu nedenle uzun ugraslar sonucu iki önemli Internet sirketi ile yaptigimiz görüsme sonucu, bize her mailde gönderecegimiz link'in tiklanmasi durumunda cocugumuzun ameliyat masraflarina 10 Cent (= 0,10 Euro) katkida bulunacaklari yönünde anlasmaya varabildik. Bu dogrultuda sizden rica edecegimiz, sizi sözkonusu sirketlerin sayfalarina yönelten bu linki tiklamaniz. Bu linki tiklayarak sizinde ameliyat masraflarini karsilamamizda katkiniz olabilir. Sizden bu zor durumumuzda istedigimiz sadece bir tik. >> TIKLAYINIZ --> http://partners.webmasterplan.com/click.asp?ref=160073&site=2514&type=text&… Bunun yaninda bu emaili tüm tanidiklariniza gönderirseniz bize son derece yardimci olursunuz. Ne kadar kisi linki tiklarsa, gerekli olan ameliyat masrafini yakalayabilmemiz o kadar olasi olur. Hos görünüz icin tesekkür eder, saygilarimi sunarim. Mehmet Can Yakupoglu.

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irg.ssu.ac.kr
by TrafficMagnet Reseller 27 Apr '03

27 Apr '03

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[Newbie]Unexpected signal from FFI on Solaris 8
by Marco Vezzoli 24 Apr '03

24 Apr '03

Hi, I'm learning how to use ffi with hugs (latest version, on Solaris 8). I can compile this simple example without errors ----Test.hs----------------------- module Test where import Foreign.C.String import Foreign.C.Types foreign import ccall "test.h incr" incr :: Int->IO Int foreign import ccall "test.h times" times :: CChar->Int->IO CString testTimes = do{j<-times (castCharToCChar 'a') 3;c <- peekCString j ;putStr c} testIncr = do{j<-incr 1;putStr $show j} ----Test.hs----------------------- ----test.c----------------------- #include <stdlib.h> #include "test.h" int incr(int i){ return i+1; } char* times(char c,int i){ printf("times running %d\n",i); char* ret; char* itr; ret=(char*)malloc(i*sizeof(char)+1); for (itr=ret;itr<ret+i;itr++){ *itr=c; } *itr=0; return ret; } ----test.c----------------------- ----test.h----------------------- int incr(int); char* times(char,int); ----test.h----------------------- the command I use is ffihugs -P{Hugs}/libraries/:{Hugs}/oldlib +G +L"test.c" Test.hs and generates Test.c and Test.so Hugs loads correctly the module but fails one of the tests: [vezzoli@web:883] ffi ->hugs -P{Hugs}/libraries/:{Hugs}/oldlib Test.hs __ __ __ __ ____ ___ _________________________________________ || || || || || || ||__ Hugs 98: Based on the Haskell 98 standard ||___|| ||__|| ||__|| __|| Copyright (c) 1994-2002 ||---|| ___|| World Wide Web: http://haskell.org/hugs || || Report bugs to: hugs-bugs(a)haskell.org || || Version: November 2002 _________________________________________ Haskell 98 mode: Restart with command line option -98 to enable extensions [loading prints removed] Test.hs Type :? for help Test> testIncr 2 Test> testTimes Unexpected signal [vezzoli@web:884] ffi -> Thank you in advance for any help. Marco -- Marco Vezzoli tel. +39 039 603 6852 STMicroelectronics fax. +39 039 603 5055

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RE: Existentials
by Hans Aberg 22 Apr '03

22 Apr '03

At 10:34 +0100 2003/04/22, Simon Marlow wrote: >If the quantifier is outside the constructor, then it scopes over all >the fields of the constructor. So, if you write > > data T = forall a . C t1 .. tn > >what should this mean (assuming you want a non-existential >interpretation)? I think you have problem here because, you apply the quantifier to something which is not a predicate. This practise makes it hard to figure out what the semantics should be. For example, in the original example: >At 09:02 +0100 2003/04/17, Simon Peyton-Jones wrote: > data T = forall a. Foo a (a -> Int) if T is a well formed formula in a quantification theory there is an associated predicate P(a) = Foo a (a -> Int) from which one gets T := all a: P(a). One of the quantification axioms that I listed said: axiom Q4. t free for x in A |- (all x: A) => [x\t]A. It means that from T, one can plug in a value t (a term) [a\t]T = [a\t]P(a) to get Foo t (t -> Int) Is that not what you want? However, in math, quantifiers are only applied to predicates, not sets. For sets, one uses operations such as union, intersection and complement. It seems me that data types correspond to sets, not predicates. Thus, one should not use quantifiers "all" and "exist" at all on them, but something corresponding to sets instead. In naive set theory, one identifies the set S := { x | P(x) } with the predicate P(x). If one should get the set S from the closed formula T above, one first applies the axiom Q4 to get P(x) := [a\x]T, and then one can get S from P(x). In the example above, the set becomes S = { a | Foo a (a -> Int) } To begin with, this does not make logically sense, as Foo a (a -> Int) is supposed to return not logical values as is required by a predicate, but data of type T. It seems me that one should instead have someting like data T = union a. Foo a (a -> Int) or data T = disjointUnion a. Foo a (a -> Int) Then, when one instantiates, one will create another addition to the data type, which is how I think is what you really want. Right? Hans Aberg

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RE: Existentials
by Simon Marlow 22 Apr '03

22 Apr '03

> "Simon Marlow" <simonmar(a)microsoft.com> writes: > > > > The example I tried failed, so I assumed it wasn't supported. > > > > You've written an existential constructor. For universal > > quantification, write it like this: > > > > data T = Foo (forall a . Enum a => a -> a) > > > > a good illustration of the confusion caused by the dual use > of forall, I > > guess :-) > > Exactly so. :-) If we change the syntax of existentials, would it > be possible to write my local universal example as I did originally > and have it work as expected? Or will it still be necessary to push > the quantifier inside the constructor? If the quantifier is outside the constructor, then it scopes over all the fields of the constructor. So, if you write data T = forall a . C t1 .. tn what should this mean (assuming you want a non-existential interpretation)? Perhaps the sensible meaning is that it is isomorphic to data T = C !(forall a . (t1,...,tn)) Cheers, Simon

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Re: Existentials
by Hans Aberg 17 Apr '03

17 Apr '03

You might determine whether it should be "all" or "exist" by means of a formal analysis: In metamathematics exist x: A is a rewriting of not(all x: not A) Then an example of a set of metamathematical axioms for the "all" quantifier is as follows (written as code in my program): theory Q. -- Quantification axioms, Mendelson, p.57 (called `K' there). { formula A, B, C term t free x. axiom Q1. A => (B => A). axiom Q2. (A => (B => C)) => ((A => B) => (A => C)). axiom Q3. (not B => not A) => ((not B => A) => B). axiom Q4. t free for x in A |- (all x: A) => [x\t]A. axiom Q5. x not free in A |- (all x: A => B) => (A => all x: B). axiom MP `modus ponens'. A, A => B |- B. axiom Gen `generalization'. A |- all x: A. } -- theory Q. Here, the symbol "|-" is the metamathematical implication, i.e., provability. Here, A |- B corresponds to Prolog B :- A. In Haskell, what is the analogue of the generalization axiom Gen? -- If you know that, then you know what should be "all" and "exist" in Haskell. Hans Aberg

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RE: Existentials
by Simon Marlow 17 Apr '03

17 Apr '03

> "Simon Peyton-Jones" <simonpj(a)microsoft.com> writes: > > > GHC and Hugs both do. In fact, GHC supports arbitrary-rank > universal > > quantification. > > The example I tried failed, so I assumed it wasn't supported. > > $ cat Exists.hs > module Exists where > data T = forall a. Enum a => Foo (a->a) > f :: T -> (Int,Bool,Char) > f (Foo g) = (g 0, g False, g 'a') > $ ghc -fglasgow-exts -c Exists.hs > > Exists.hs:4: > Couldn't match `Int' against `Bool' > Expected type: Int > Inferred type: Bool > In the first argument of `g', namely `False' > In the definition of `f': (g 0, g False, g 'a') You've written an existential constructor. For universal quantification, write it like this: data T = Foo (forall a . Enum a => a -> a) a good illustration of the confusion caused by the dual use of forall, I guess :-) Cheers, Simon

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