Peter Cai:
Hi all,
In order to improve my understanding of monad, I am trying to do some manual computation on "Reader Monad" but I got some problem.
The computation is like this:
--instance Monad (Reader e) where -- return a = Reader $ \e -> a -- (Reader r) >>= f = Reader $ \e -> f (r e) e
This cannot work out because this definition is ill-typed, which you will notice if you try to compile it (or if you contemplate about what the type of (f (r e) e) is). Here is a way to define a reader monad: newtype Reader e a = Reader { runReader :: (e -> a) } instance Monad (Reader e) where return a = Reader $ \_ -> a Reader r >>= f = Reader $ \e -> runReader (f (r e)) e You can then do your calculation like this: Reader show >>= return = Reader $ \e -> runReader (return (show e)) e = Reader $ \e -> runReader (Reader (\_ -> show e)) e = Reader $ \e -> (\_ -> show e) e = Reader $ \e -> show e = Reader show; -- Note that, except for the use of “show” as a specific -- example, the above is equivalent to proving that the -- second monad law holds for (Reader e). runReader (Reader show >>= return) = runReader (Reader show) = show. Malte