On Sat, Oct 17, 2009 at 3:24 PM, Andrew Coppin wrote:

Edward Z. Yang wrote:

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Excerpts from Andrew Coppin's message of Sat Oct 17 15:21:28 -0400 2009:

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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)

In this case it is trivial, Foo = ReaderT M IO which is a monad. If you want, you can verify for yourself that this is a monad. It isn't, however, uncommon for custom monads to be (equivalent to) stacks of monad transformers.

Am Samstag 17 Oktober 2009 22:24:08 schrieb Andrew Coppin:

Edward Z. Yang wrote:

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Excerpts from Andrew Coppin's message of Sat Oct 17 15:21:28 -0400 2009:

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Suppose we have

newtype Foo x instance Monad Foo runFoo :: Foo x -> IO x

Are you sure that's really the type you want for runFoo ? By the definition below, runFoo :: Foo x -> M -> IO x would be the natural type.

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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

So, (Foo f) >>= return ~> Foo $ \m -> do x <- f m let Foo g = return x g m ~> (since g is (\_ -> return x)) Foo $ \m -> do x <- f m (\_ -> return x) m ~> Foo $ \m -> do x <- f m return x ~> (monad laws for IO) Foo $ \m -> f m ~> Foo f That's number 1. return a >>= f ~> Foo $ \m -> do x <- (\_ -> return a) m let Foo g = f x g m ~> Foo $ \m -> do x <- return a let Foo g = f x g m ~> Foo $ \m -> do let Foo g = f a g m ~> Foo $ \m -> unFoo (f a) m ~> Foo (unFoo $ f a) ~> f a where I've used unFoo :: Foo x -> M -> IO x; unFoo (Foo f) = f, that's number 2. (Foo f >>= g) >>= h ~> Foo $ \m -> do y <- unFoo (Foo f >>= g) m let Foo h1 = h y h1 m ~> Foo $ \m -> do y <- (unFoo . Foo $ \m1 -> do x <- f m1 let Foo g1 = g x g1 m1) m let Foo h1 = h y h1 m ~> Foo $ \m -> do y <- do x <- f m let Foo g1 = g x g1 m let Foo h1 = h y h1 m ~> Foo $ \m -> do y <- f m >>= \x -> unFoo (g x) m unFoo (h y) m ~> Foo $ \m -> (f m >>= \x -> unFoo (g x) m) >>= \y -> unFoo (h y) m ~> Foo $ \m -> f m >>= \x -> (unFoo (g x) m >>= \y -> unFoo (h y) m) And Foo f >>= (\z -> g z >>= h) ~> Foo $ \m -> do x <- f m let Foo k = (\z -> g z >>= h) x k m ~> Foo $ \m -> do x <- f m unFoo (g x >>= h) m ~> Foo $ \m -> f m >>= \x -> unFoo (g x >>= h) m ~> Foo $ \m -> f m >>= \x -> (\m1 -> (unFoo (g x) m1) >>= \y -> unFoo (h y) m1) m ~> Foo $ \m -> f m >>= \x -> (unFoo (g x) m >>= \y -> unFoo (h y) m) That's number 3.

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)"...