On November 18, 2015 12:15:37 AM GMT+02:00, Marcin Mrotek
Hello,
I'm pretty sure a Pair (like any other fixed-length type, besides the corner case of a single field like in Writer, Identity, etc) can't be a monad. Perhaps instead of struggling with >>=, consider join. It has a type:
join :: Monad m => m (m a) -> m a
for Pairs that would be
join :: Pair (Pair a) -> Pair a join (Pair (Pair a1 a2) (Pair b1 b2)) = Pair _ _
How do you want to fit four values into two boxes? You cannot place any constraints on the type inside the pair, so it can't be a monoid or anything that would let you combine the values somehow. You could only choose two of the values and drop the other two on the floor.
Getting back to >>=, it's assumed to follow these laws:
1) return a >>= k = k a 2) m >>= return = m 3) m >>= (\x -> k x >>= h) = (m >>= k) >>= h
As for the firs, return a = Pair a a. Then the first two laws become
1) Pair a a >>= k = k a 2) Pair a b >>= (\a -> Pair a a) = Pair a b
The first law could work if >>= just chose one of the values arbitrarily. But the second law is a hopeless case. You would need to pick one element of a pair, plug it into a function that repeats the argument, and somehow get back the other element that you've already dropped.
Concluding, either I'm sorely mistaken or there indeed isn't a Monad instance for Pair.
Best regards, Marcin Mrotek _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
No. Pair can be made into a monad, and this is made clear by realizing that Pair a ~ Bool -> a, so the monad instance for Reader should work here. In fact, this means that any representable functor (i.e. f s.t. f a ~ t-> a for some t) has all instances that Reader r has for fixed r. HTH, Gesh