Looks to me as if you have OverloadedStrings enabled somewhere, in which case that would be the correct behaviour. IRS ________________________________________ From: haskell-cafe-bounces@haskell.org [haskell-cafe-bounces@haskell.org] on behalf of Vlatko Basic [vlatko.basic@gmail.com] Sent: Wednesday, June 26, 2013 11:05 AM Cc: Haskell-Cafe Subject: [Haskell-cafe] Annotation problem after HP reinstalation I uninstalled haskell-platform, deleted .cabal and .ghc dirs and reinstalled it (on Ubuntu), but now the annotation (and other annotations) {-# ANN module "HLint: ignore Eta reduce" #-} produces error Ambiguous type variable `a0' in the constraints: (Data a0) arising from an annotation at src/Model.hs:82:1-45 (Data.String.IsString a0) arising from the literal `"HLint: ignore Eta reduce"' at src/Model.hs:82:16-41 Probable fix: add a type signature that fixes these type variable(s) In the expression: "HLint: ignore Eta reduce" In the annotation: {-# ANN module "HLint: ignore Eta reduce" #-} Have I missed to install/configure something? vlatko _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

Hello Cafe! I had a (simplified) record data P = P { a :: String, b :: String, c :: IO String } deriving (Show, Eq) but to get automatic deriving of 'Show' and 'Eq' for 'data P' I have created 'newtype IOS' and its 'Show' and 'Eq' instances newtype IOS = IO String instance Show (IOS) where show _ = "(IO String) function" instance Eq (IOS) where _ == _ = True (the easiest 'instance Show (IO String) where ...' produces "orphan instance") and changed 'data P' to data P = P { a :: String, b :: String, c :: IOS } deriving (Show, Eq) but now when I try to set 'c' field in fun :: FilePath -> P -> IO P fun path p = do b <- doesFileExist path ... return $ p {c = readFile path} I get error Couldn't match expected type `IOS' with actual type `IO String' which is correct. So, the question is: Is it possible to somehow "cast" 'IO String' to 'IOS', or in some other way set 'c' field to an 'IO String' function so that it can be later used such as 'content <- c p'. Thanks

On 1 Jul 2013, at 16:07, Vlatko Basic wrote:

I had a (simplified) record

data P = P { a :: String, b :: String, c :: IO String } deriving (Show, Eq)

but to get automatic deriving of 'Show' and 'Eq' for 'data P' I have created 'newtype IOS' and its 'Show' and 'Eq' instances

newtype IOS = IO String

Not quite! That is a newtype'd String, not a newtype's (IO String). Try this: newtype IOS = IOS (IO String)

but now when I try to set 'c' field in

return $ p {c = readFile path}

I get error Couldn't match expected type `IOS' with actual type `IO String'

Use the newtype constructor to convert an IO String -> IOS. return $ p {c = IOS $ readFile path} Regards, Malcolm

Is there a nicer way to extract the 'IO String' from 'IOS', without 'case' or without pattern matching the whole 'P'? newtype IOS = IOS (IO String) data P = P { getA :: String, getB :: String, getC :: IOS } deriving (Show, Eq) getC_IO :: P -> IO String getC_IO p = case getC p of IOS a -> a getC_IO (P _ _ (IOS a)) = a -------- Original Message -------- Subject: Re: [Haskell-cafe] "Casting" newtype to base type? From: Malcolm Wallace To: vlatko.basic@gmail.com Cc: Haskell-Cafe Date: 01.07.2013 17:24

On 1 Jul 2013, at 16:07, Vlatko Basic wrote:

...

I had a (simplified) record

data P = P { a :: String, b :: String, c :: IO String } deriving (Show, Eq)

but to get automatic deriving of 'Show' and 'Eq' for 'data P' I have created 'newtype IOS' and its 'Show' and 'Eq' instances

newtype IOS = IO String

Not quite! That is a newtype'd String, not a newtype's (IO String). Try this:

newtype IOS = IOS (IO String)

...

but now when I try to set 'c' field in

return $ p {c = readFile path}

I get error Couldn't match expected type `IOS' with actual type `IO String'

Use the newtype constructor to convert an IO String -> IOS.

return $ p {c = IOS $ readFile path}

Regards, Malcolm

-------- Original Message -------- Subject: Re: [Haskell-cafe] "Casting" newtype to base type? From: Tom Ellis To: haskell-cafe@haskell.org Date: 02.07.2013 15:25

On Tue, Jul 02, 2013 at 03:03:08PM +0200, Vlatko Basic wrote:

...

Is there a nicer way to extract the 'IO String' from 'IOS', without 'case' or without pattern matching the whole 'P'?

newtype IOS = IOS (IO String) data P = P { getA :: String, getB :: String, getC :: IOS } deriving (Show, Eq)

getC_IO :: P -> IO String getC_IO p = case getC p of IOS a -> a getC_IO (P _ _ (IOS a)) = a

How about

unIOS :: IOS -> IO String unIOS (IOS a) = a

getC_IO :: P -> IO String getC_IO = unIOS . getC

Thanks for your answer. I had those two funcs, but thought there might be a shorter/prettier one-func one-liner. :-)

Tom

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Hi Vlatko. On 2 July 2013 16:03, Vlatko Basic wrote:

Is there a nicer way to extract the 'IO String' from 'IOS', without 'case' or without pattern matching the whole 'P'?

You might enjoy the newtype package. http://hackage.haskell.org/package/newtype Hope this helps, Ozgur.

I'm experimenting. The IO field is just a helper field, so it shouldn't have any consequences. -------- Original Message -------- Subject: Re: [Haskell-cafe] "Casting" newtype to base type? From: Tom Ellis To: haskell-cafe@haskell.org Date: 01.07.2013 17:24

On Mon, Jul 01, 2013 at 05:07:00PM +0200, Vlatko Basic wrote:

...

Hello Cafe!

I had a (simplified) record

data P = P { a :: String, b :: String, c :: IO String } deriving (Show, Eq)

but to get automatic deriving of 'Show' and 'Eq' for 'data P' I have created 'newtype IOS' and its 'Show' and 'Eq' instances

newtype IOS = IO String instance Show (IOS) where show _ = "(IO String) function" instance Eq (IOS) where _ == _ = True

An Eq instance for something containing IO is bound to lead to puzzlement somewhere down the line. I think you're better off defining something like

data P_lesser = P_lesser { a_lesser :: String, b_lesser :: String } deriving (Show, Eq)

to_lesser p = P_lesser (a p) (b p)

and just factoring everything through "to_lesser" when you want to compare or show.

Tom

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