[Haskell-cafe] How to "instance MonadIO Identity"?
magicloud.magiclouds at gmail.com
Tue Dec 28 07:18:36 CET 2010
Wow, this explains a lot. Thanks.
On Tue, Dec 28, 2010 at 2:05 PM, Antoine Latter <aslatter at gmail.com> 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
> 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.
> On Tue, Dec 28, 2010 at 12:47 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).
>>> On Tue, Dec 28, 2010 at 7:29 AM, Magicloud Magiclouds
>>> <magicloud.magiclouds at gmail.com> wrote:
>>>> From another thread in this list, I got code as:
>>>>> instance MonadIO Identity where
>>>>> liftIO = id
>>>> Well, it does not work for me as:
>>>> 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
>>>> Haskell-Cafe mailing list
>>>> Haskell-Cafe at haskell.org
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
More information about the Haskell-Cafe