Edward Z. Yang wrote:
Excerpts from Andrew Coppin's message of Sat Oct 17 15:21:28 -0400 2009:
Suppose we have
newtype Foo x instance Monad Foo runFoo :: Foo x -> IO x
What sort of things can I do to check that I actually implemented this correctly? I mean, ignoring what makes Foo special for a moment, how can I check that it works correctly as a monad.
A proper monad obeys the monad laws:
http://www.haskell.org/haskellwiki/Monad_Laws
You can probably cook up some quickcheck properties to test for these, but really you should be able to convince yourself by inspection that your monad follows these laws.
I'm reasonably confident it works, but not 100% sure... newtype Foo x = Foo (M -> IO x) instance Monad Foo where return x = Foo (\_ -> return x) (Foo f1) >>= fn = Foo $ \m -> do x <- f1m let Foo f2 = fn x f2 m runFoo (Foo f) = do let m = ... f m I can do something like "runFoo (return 1)" and check that it yields 1. (If it doesn't, my implementation is *completely* broken!) But I'm not sure how to proceed further. I need to assure myself of 3 things: 1. runFoo correctly runs the monad and retrives its result. 2. return does that it's supposed to. 3. (>>=) works correctly. Doing "runFoo (return 1)" seems to cover #1 and #2, but I'm not so sure about #3... I guess I could maybe try "runFoo (return 1 >>= return)"...