[Haskell-cafe] Multi-Class monadic type?

[Alexander Treptow]

Edit:
it is running,
testFunc :: (forall a m. (Conf a, MonadIO m) => m a) -> TestType
is the solution ;)

----------
Thanks,
i tried it with your code changes but then i ll only get a type mismatch

Test.hs:12:11:
Couldn't match expected type `forall a. (Conf a, MonadIO m) => m a'
       against inferred type `forall a (m1 :: * -> *). (Conf a, MonadIO 
m1) => m1 a'
In the expression: TestType
In the definition of `testFunc': testFunc = TestType


Miguel Mitrofanov schrieb:
>/ Error message suggests that you've used "Conf" improperly.
/>/
/>/ testFunc :: (forall a. Conf a, MonadIO m => m a) -> TestType
/>/
/>/ is illegal, as I recall, you should use another pair of brackets:
/>/
/>/ testFunc :: (forall a. (Conf a, MonadIO m) => m a) -> TestType
/>/
/>/ Alexander Treptow wrote:
/>>/ Hi,
/>>/ i got a little problem and don't know how to solve that. Hope you can
/>>/ help me.
/>>/
/>>/ code:
/>>/ ----------
/>>/ module Test where
/>>/ {-# LANGUAGE Rank2Types, RankNTypes #-}
/>>/ import Control.Monad.Trans
/>>/
/>>/ data TestType = TestType
/>>/  {tst :: (Conf a, MonadIO m) => m a}
/>>/
/>>/ class Conf a where
/>>/  get :: MonadIO m => m a
/>>/
/>>/ testFunc :: (forall a. Conf a, MonadIO m => m a) -> TestType
/>>/ testFunc = TestType
/>>/ --------------
/>>/ error:
/>>/ --------------
/>>/ Test.hs:11:23
/>>/    Class `Conf' used as a type
/>>/    In the type signature for `testFunc':
/>>/      testFunc :: (forall a. Conf a, (MonadIO m) => m a) -> TestType
/>>/ --------------
/>>/
/>>/ explanation:
/>>/ I need a data type that creates a record with a member that has no fixed
/>>/ type, because its not known at compile-time. The Rank2Types language
/>>/ extension fits that need, but i figured out that i ll need to make that
/>>/ time monadic to avoid the use of unsafePerformIO in the program that
/>>/ uses this lib.
/>>/
/>>/ Thanks and greetings,
/>>/    Alex
/>>/
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/>>/
/>/
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