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

[Michael Snoyman]

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
<magicloud.magiclouds at gmail.com> 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 <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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>>
>
>
>
> --
> 竹密岂妨流水过
> 山高哪阻野云飞
>