[Haskell-cafe] Problem with forall type in type declaration

[Yves Parès]

run :: Monad IO a -> IO a

Actually this type is wrong. Monad has to appear as a class constraint, for
instance :

run :: Monad m => m a -> IO a

Are you trying to make:

run :: IO a -> IO a
??

2012/5/4 Magicloud Magiclouds <magicloud.magiclouds at gmail.com>

> Hi,
>  Assuming this:
> run :: Monad IO a -> IO a
> data Test = Test { f }
>
>  Here I'd like to set f to run, like "Test run". Then what is the type of
> f?
>  The confusing (me) part is that, the argument pass to f is not fixed
> on return type, like "f1 :: Monad IO ()", "f2 :: Monad IO Int". So
> "data Test a = Test { f :: Monad IO a -> IO a} does not work.
> --
> 竹密岂妨流水过
> 山高哪阻野云飞
>
> And for G+, please use magiclouds#gmail.com.
>
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