[Haskell-cafe] How to "instance MonadIO Identity"?

[Magicloud Magiclouds]

  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 <michael at snoyman.com>:
> 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
> <magicloud.magiclouds at gmail.com> 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
>>
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>



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竹密岂妨流水过
山高哪阻野云飞