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:

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Hi,  From another thread in this list, I got code as:

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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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