How to "instance MonadIO Identity"?

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Magicloud Magiclouds

28 Dec 2010 28 Dec '10

12:29 a.m.

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

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Michael Snoyman

28 Dec 28 Dec

12:32 a.m.

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

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

Michael Snoyman

12:56 a.m.

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

Antoine Latter

1:05 a.m.

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

Magicloud Magiclouds

1:18 a.m.

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

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Antoine Latter
Magicloud Magiclouds
Michael Snoyman