Hi In future, you are probably best off addressing these general questions to the haskell-cafe@ mailing list.
> f a = let m = id a in do return m We can simplify this to: f a = let m = id a in do return m f a = let m = a in do return m f a = do return a f a = return a This shows why f has the same type as return. > f' a = do m <- id a > return m f' a = id a >>= \m -> return m f' a = a >>= \m -> return m You basically return a, but have sent it through >>= which means it must be monadic. > g a = let m =[a,a] in do return m g a = return [a,a] > g' a = do m <- [a,a] > return m This is the tricky one. Here your use of m <- [a,a] has bound the Monad in question to be the list Monad, hence you get a result in terms of lists, not monads. do notation applies to any monad, including list. Thanks Neil _______________________________________________ Hugs-Users mailing list Hugs-Users@haskell.org http://www.haskell.org/mailman/listinfo/hugs-users
