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

28 Dec 2010


      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
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