On Fri, 2008-10-31 at 18:43 -0200, Mauricio wrote:

Hi,

After a lot of thinking, I can't get what I am doing wrong in this code:

------ data ( RandomGen g ) => RandomMonad g a = RandomMonad (g -> a)

instance Monad (RandomMonad g) where return = RandomMonad . const RandomMonad f1 >>= f2 = RandomMonad f3 where f3 a = f2f1 a (next a) RandomMonad f2f1 = f2 . f1 ------

I get this error message:

Could not deduce (RandomGen g) from the context (Monad (RandomMonad g)) arising from a use of `RandomMonad' at src/encherDB.hs:10:11-21 Possible fix: add (RandomGen g) to the context of the type signature for `return' In the first argument of `(.)', namely `RandomMonad' In the expression: RandomMonad . const In the definition of `return': return = RandomMonad . const

but I'm not smart enough to understand what it means.

Yikes. Just search the archives for `Set not a Monad'; you have the same issue. Hint: data RandomGen g => RandomMonad g a means nothing at all like what you think it means. jcc

On Fri, 31 Oct 2008, Mauricio wrote:

Hi,

After a lot of thinking, I can't get what I am doing wrong in this code:

------ data ( RandomGen g ) => RandomMonad g a = RandomMonad (g -> a)

RandomGen g is considered the constraint for the application of RandomMonad constructor, but GHC does not conclude that every value of (RandomMonad g a) fulfills this constraint. Actually, 'undefined' is available for any 'g'.

instance Monad (RandomMonad g) where return = RandomMonad . const RandomMonad f1 >>= f2 = RandomMonad f3 where f3 a = f2f1 a (next a) RandomMonad f2f1 = f2 . f1

you need to make (RandomGen g) a constraint of the instance: instance RandomGen g => Monad (RandomMonad g) where Btw. RandomMonad looks like Control.Monad.Reader.