[Haskell-cafe] Newbie on instance of Monad

[Jonathan Cast]

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