I think of (r -> m a) as a type signature and Int or Bool by themselves as types. So, all type signatures are themselves types?
Michael
--- On Wed, 12/29/10, Henning Thielemann
In the case of ReaderT and StateT
newtype ReaderT r m a = ReaderT { -- | The underlying computation, as a function of the environment. runReaderT :: r -> m a }
newtype StateT s m a = StateT { runStateT :: s -> m (a, s) }
what is the existing type?
The existing type is 'r -> m a'. You could also write
newtype ReaderT r m a = ReaderT (r -> m a)
This would be the same type as above, but it would have no accessor function 'runReaderT'.