Re: [Haskell-cafe] How to "instance MonadIO Identity"?
Wow, this explains a lot. Thanks. On Tue, Dec 28, 2010 at 2:05 PM, Antoine Latter wrote: > There are a couple of ways to handle this - either call > getSystemEncoding on the outside, before calling 'runWriter', and then > pass it in to your writer computation, or use 'WriterT IO' instead of > 'Writer'. > > That would be something like: > > main = do > result <- runWriterT ( computation involving 'tell' and 'liftIO') > ( computation using result) > > The error message in your case can be difficult to understand - the > 'Writer' type is a synonym for the type 'WriterT Identity'. Since none > of these types involve IO, then liftIO can't be used. > > Antoine > > > On Tue, Dec 28, 2010 at 12:47 AM, Magicloud Magiclouds > wrote: >> Ah, that is a bad news. >> I am using Control.Monad.Writer and Data.Encoding. Code like (not >> compilable) >> 43|instance WithMessage String where >> 44| append s = (liftIO $ getSystemEncoding) >>= (\e -> tell $ >> encodeLazyByteString e s) >> May I know how to make this work? >> >> 2010/12/28 Michael Snoyman : >>> The only way to create such an instance would be with unsafePerformIO, >>> which in this case would be a Very Bad Idea (tm). >>> >>> Michael >>> >>> On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds >>> wrote: Hi, From another thread in this list, I got code as: > instance MonadIO Identity where > liftIO = id Well, it does not work for me as: Message.hs:22:12: Couldn't match expected type `Identity a' with actual type `IO a' Expected type: IO a -> Identity a Actual type: IO a -> IO a In the expression: id In an equation for `liftIO': liftIO = id -- 竹密岂妨流水过 山高哪阻野云飞 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe >>> >> >> >> >> -- >> 竹密岂妨流水过 >> 山高哪阻野云飞 >> >> ___ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> > -- 竹密岂妨流水过 山高哪阻野云飞 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] How to "instance MonadIO Identity"?
There are a couple of ways to handle this - either call getSystemEncoding on the outside, before calling 'runWriter', and then pass it in to your writer computation, or use 'WriterT IO' instead of 'Writer'. That would be something like: main = do result <- runWriterT ( computation involving 'tell' and 'liftIO') ( computation using result) The error message in your case can be difficult to understand - the 'Writer' type is a synonym for the type 'WriterT Identity'. Since none of these types involve IO, then liftIO can't be used. Antoine On Tue, Dec 28, 2010 at 12:47 AM, Magicloud Magiclouds wrote: > Ah, that is a bad news. > I am using Control.Monad.Writer and Data.Encoding. Code like (not compilable) > 43|instance WithMessage String where > 44| append s = (liftIO $ getSystemEncoding) >>= (\e -> tell $ > encodeLazyByteString e s) > May I know how to make this work? > > 2010/12/28 Michael Snoyman : >> The only way to create such an instance would be with unsafePerformIO, >> which in this case would be a Very Bad Idea (tm). >> >> Michael >> >> On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds >> wrote: >>> Hi, >>> From another thread in this list, I got code as: instance MonadIO Identity where liftIO = id >>> Well, it does not work for me as: >>> Message.hs:22:12: >>> Couldn't match expected type `Identity a' with actual type `IO a' >>> Expected type: IO a -> Identity a >>> Actual type: IO a -> IO a >>> In the expression: id >>> In an equation for `liftIO': liftIO = id >>> -- >>> 竹密岂妨流水过 >>> 山高哪阻野云飞 >>> >>> ___ >>> Haskell-Cafe mailing list >>> Haskell-Cafe@haskell.org >>> http://www.haskell.org/mailman/listinfo/haskell-cafe >>> >> > > > > -- > 竹密岂妨流水过 > 山高哪阻野云飞 > > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] How to "instance MonadIO Identity"?
Well, you *could* use unsafePerformIO on getSystemEncoding, though I'm not familiar with getSystemEncoding, so I can't tell you whether this is an acceptable usage of unsafePerformIO. On Tue, Dec 28, 2010 at 7:47 AM, Magicloud Magiclouds wrote: > Ah, that is a bad news. > I am using Control.Monad.Writer and Data.Encoding. Code like (not compilable) > 43|instance WithMessage String where > 44| append s = (liftIO $ getSystemEncoding) >>= (\e -> tell $ > encodeLazyByteString e s) > May I know how to make this work? > > 2010/12/28 Michael Snoyman : >> The only way to create such an instance would be with unsafePerformIO, >> which in this case would be a Very Bad Idea (tm). >> >> Michael >> >> On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds >> wrote: >>> Hi, >>> From another thread in this list, I got code as: instance MonadIO Identity where liftIO = id >>> Well, it does not work for me as: >>> Message.hs:22:12: >>>Couldn't match expected type `Identity a' with actual type `IO a' >>>Expected type: IO a -> Identity a >>> Actual type: IO a -> IO a >>>In the expression: id >>>In an equation for `liftIO': liftIO = id >>> -- >>> 竹密岂妨流水过 >>> 山高哪阻野云飞 >>> >>> ___ >>> Haskell-Cafe mailing list >>> Haskell-Cafe@haskell.org >>> http://www.haskell.org/mailman/listinfo/haskell-cafe >>> >> > > > > -- > 竹密岂妨流水过 > 山高哪阻野云飞 > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] How to "instance MonadIO Identity"?
Ah, that is a bad news. I am using Control.Monad.Writer and Data.Encoding. Code like (not compilable) 43|instance WithMessage String where 44| append s = (liftIO $ getSystemEncoding) >>= (\e -> tell $ encodeLazyByteString e s) May I know how to make this work? 2010/12/28 Michael Snoyman : > The only way to create such an instance would be with unsafePerformIO, > which in this case would be a Very Bad Idea (tm). > > Michael > > On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds > wrote: >> Hi, >> From another thread in this list, I got code as: >>> instance MonadIO Identity where >>> liftIO = id >> Well, it does not work for me as: >> Message.hs:22:12: >> Couldn't match expected type `Identity a' with actual type `IO a' >> Expected type: IO a -> Identity a >> Actual type: IO a -> IO a >> In the expression: id >> In an equation for `liftIO': liftIO = id >> -- >> 竹密岂妨流水过 >> 山高哪阻野云飞 >> >> ___ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe >> > -- 竹密岂妨流水过 山高哪阻野云飞 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] How to "instance MonadIO Identity"?
The only way to create such an instance would be with unsafePerformIO, which in this case would be a Very Bad Idea (tm). Michael On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds wrote: > Hi, > From another thread in this list, I got code as: >> instance MonadIO Identity where >> liftIO = id > Well, it does not work for me as: > Message.hs:22:12: >Couldn't match expected type `Identity a' with actual type `IO a' >Expected type: IO a -> Identity a > Actual type: IO a -> IO a >In the expression: id >In an equation for `liftIO': liftIO = id > -- > 竹密岂妨流水过 > 山高哪阻野云飞 > > ___ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] How to "instance MonadIO Identity"?
Hi, From another thread in this list, I got code as: > instance MonadIO Identity where > liftIO = id Well, it does not work for me as: Message.hs:22:12: Couldn't match expected type `Identity a' with actual type `IO a' Expected type: IO a -> Identity a Actual type: IO a -> IO a In the expression: id In an equation for `liftIO': liftIO = id -- 竹密岂妨流水过 山高哪阻野云飞 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
