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Re: Q: Resolving ambiguous type variable
by tty＠SAFe-mail.net 17 May '02

17 May '02

Would it be correct to conclude from your reasoning that the following would be properly typed by the compiler ? first_h :: (Show a) => [a] -> Result a first_h [] = Ex "No list" first_h (x:xs) = Value x since we are explicitly stating first_h can only take a list containing elements deriving Show. My expectations would be that this would be unambiguous, unfortunately ghc-5.02.3 still complains with the reported error. Regards Tee -------- Original Message -------- From: hdaume(a)ISI.EDU To: tty(a)SAFe-mail.net Cc: haskell-cafe(a)haskell.org Subject: Re: Q: Resolving ambiguous type variable Date: Fri, 17 May 2002 10:24:29 -0700 (PDT) > Erm, I think I said something stupid (or at least implied it). > > main = getArgs >>= print . first_h > > *will* work, because it knows that the result of a getArgs is a list of > Strings. > > The problem you originally had was that when you say: > > first_h [] > > The "[]" could refer to a list of any type of element. Since not every > element type is an instance of Show, it doesn't know that it can apply > show to the list. > > By saying, for instance, (first_h ([] :: [Char])) you're saying "okay, > this is an empty list, but it's a list of Characters and you know that a > character can be shown." If you backtrack what the type checker is doing, > it says: > > print (first_h []) > > okay, the argument to print must be an instance of Show (print has type > Show a => a -> IO ()). therefore, > > first_h [] > > must be an instance of Show. well, what's the type of first_h []? Well, > the type of first_h is [a] -> Result a. is "Result a" an instance of > show? well, let's see...you said data Result a = ... deriving > (Show). this basically means you get an instance declaration: > > instance Show a => Show (Result a) where > ...some stuf... > > so, we want to know if Result a is an instance of show. Well, it is > whenever a is an instance of show. so, if the elements of your list are > an instance of show, then (first_h []) is an instance of Show, as we > require. > > however, since you just say "[]" it knows this is of type [a], but it > *doesn't* know what a is. therefore, a could either be or not be an > instance of show. however, if you explicitly specify what a is and that > explicit type is an instance of show (like Char in my example), then > everything is fine. > > -- > Hal Daume III > > "Computer science is no more about computers | hdaume(a)isi.edu > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > On Fri, 17 May 2002, Hal Daume III wrote: > > > In short, you cannot. > > > > What if your main were: > > > > main = getArgs >>= print . first_h > > > > The compiler doesn't know the difference and so it needs a type. > > > > Simple fix: > > > > > main = print (first_h ([] :: [Char])) > > > > > > -- > > Hal Daume III > > > > "Computer science is no more about computers | hdaume(a)isi.edu > > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > > > On Fri, 17 May 2002 tty(a)SAFe-mail.net wrote: > > > > > Hello, > > > I am writing a function which returns an exception/error string should an unexpected parameter is passed in. Borrowing from the Maybe module I have the following: > > > ------------------------------------------------------------ > > > data Result a = Ex String | Value a deriving Show > > > > > > -- Testing the Result a type > > > first_h :: [a] -> Result a > > > first_h [] = Ex "No list" > > > first_h (x:xs) = Value x > > > > > > -- Trying out the first_h > > > main :: IO() > > > main = print (first_h []) > > > ------------------------------------------------------------ > > > > > > Which the compiler complains: > > > > > > Ambiguous type variable(s) `a' in the constraint `Show a' > > > arising from use of `print' at atype.hs:8 > > > In the definition of `main': print (first_h []) > > > > > > This is understandable since it is unable to unify the empty list with a concrete list of type 'a' i.e. there are an infinite types which would match. My question is how can I indicate to the compiler that it does not matter which type the empty list is since the return result is of a concreate type. > > > > > > Thanks > > > > > > Tee > > > _______________________________________________ > > > Haskell-Cafe mailing list > > > Haskell-Cafe(a)haskell.org > > > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > > > _______________________________________________ > > Haskell-Cafe mailing list > > Haskell-Cafe(a)haskell.org > > http://www.haskell.org/mailman/listinfo/haskell-cafe > >

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Q: Resolving ambiguous type variable
by tty＠SAFe-mail.net 17 May '02

17 May '02

Hello, I am writing a function which returns an exception/error string should an unexpected parameter is passed in. Borrowing from the Maybe module I have the following: ------------------------------------------------------------ data Result a = Ex String | Value a deriving Show -- Testing the Result a type first_h :: [a] -> Result a first_h [] = Ex "No list" first_h (x:xs) = Value x -- Trying out the first_h main :: IO() main = print (first_h []) ------------------------------------------------------------ Which the compiler complains: Ambiguous type variable(s) `a' in the constraint `Show a' arising from use of `print' at atype.hs:8 In the definition of `main': print (first_h []) This is understandable since it is unable to unify the empty list with a concrete list of type 'a' i.e. there are an infinite types which would match. My question is how can I indicate to the compiler that it does not matter which type the empty list is since the return result is of a concreate type. Thanks Tee

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by tony 17 May '02

17 May '02

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Simple Question
by Jerry, JiJie 16 May '02

16 May '02

Hi, enlightenment much appreciated for my newbie question -- to achieve: append' [ [1, 2], [3, 4], [5] ] 6 -> [ [1, 2], [3, 4], [5, 6] ] append' [ ['1', '2'], ['3'] ] '4' -> [ ['1', '2'], ['3', '4'] ] append' [ [True], [True] ] False -> [ [True], [True, False] ] so I (naively) write: 18) append' :: [[a]] -> a -> [[a]] 19) append' [] y = [[y]] 20) append' x:xs y = [(init x:xs)] ++ [(tail xs)++[y]] and ghc(5.02.3) says: $ ghc p3a.hs p3a.hs:20: Parse error in pattern thanx in advance! -- Regards, Jerry

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by raeline 16 May '02

16 May '02

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��ȫ��ͼ�
by raeline 15 May '02

15 May '02

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RE: Replacing the Prelude
by Simon Peyton-Jones 14 May '02

14 May '02

Ashley writes | > I was hoping to do something similar for 'do' notation by redefining | > (>>), (>>=) etc., but unfortunately GHC is quite insistent | that 'do' notation quite specifically refers to GHC.Base.Monad Dylan replies | I'm surprised that ghc uses the fromInteger and fromRational | that are in scope; I thought there was general agreement that | it should use the Prelude.from{Integer,Rational} in scope. Ashley is referring to a GHC extension. Normally, GHC uses Prelude.fromInteger etc, regardless of what is in scope. But if you say -fno-implicit-prelude, GHC will instead use whatver fromInteger is in scope. (This is documented in the manual.) Ashley's question, as I understand is whether something similar could be done for monads. Answer: I think so, without much bother. I'm beavering away on a Haskell workshop paper at the moment, but ping me in a fortnight to do it. Simon

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Re: Replacing the Prelude
by Dylan Thurston 14 May '02

14 May '02

On Sun, May 12, 2002 at 09:31:38PM -0700, Ashley Yakeley wrote: > I have recently been experimenting writing code that replaces large > chunks of the Prelude, compiling with -fno-implicit-prelude. I notice > that I can happily redefine numeric literals simply by creating functions > called 'fromInteger' and 'fromRational': GHC will use whatever is in > scope for those names. > > I was hoping to do something similar for 'do' notation by redefining > (>>), (>>=) etc., but unfortunately GHC is quite insistent that 'do' > notation quite specifically refers to GHC.Base.Monad (i.e. Prelude.Monad, > as the Report seems to require). I don't suppose there's any way of > fooling it, is there? I was rather hoping 'do' notation would work like a > macro in rewriting its block, and not worry about types at all. > > I accept that this might be a slightly bizarre request. There are a > number of things I don't like about the way the Prelude.Monad class and > 'do' notation are set up, and it would be nice to be able to experiment > with alternatives. A while ago, there were extensive discussions about replacing the Prelude on this list. (Search for "Prelude shenanigans".) I started to write up a design document for how to enable replacing the Prelude. This boiled down to taking most of the "syntactic sugar" defined in the report seriously, ignoring the types (as you say). I'm surprised that ghc uses the fromInteger and fromRational that are in scope; I thought there was general agreement that it should use the Prelude.from{Integer,Rational} in scope. As I recall, there were several relevant bits of syntactic sugar: - Numeric types, including 'fromInteger', 'fromRational', and 'negate'. This all works fine, except that the defaulting mechanism is completely broken, causing a number of headaches. - Monads. The translation given in the report is clean, and it seems like it can be used without problems. - Bools. There was a slight problem here: the expansion of 'if ... then ... else ...' uses a case construct referring to the constructors of the Bool type, which prevents any useful redefinition of Bool. I would propose using a new function, 'Prelude.ifThenElse', if there is one in scope. - Enumerations. Clean syntactic sugar. - List comprehensions. The report gives one translation, but I think I might prefer a translation into monads. - Lists and tuples more generally. At some point the translations start getting too hairy; I think I decided that lists and tuples were too deeply intertwined into the language to change cleanly. I'll dig up my old notes and write more, and then maybe write a complete design document and get someone to implement it. --Dylan Thurston

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composeList
by Emre Tezel 12 May '02

12 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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12 May '02

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