10 Mar
2008
10 Mar
'08
1:10 p.m.
Neil Mitchell wrote:
instance Eq Foo where (==) (Foo a _) (Foo b _) = (==) a b [...] Please give the sane law that this ordering violates. I can't see any!
The (non-existant) law would be (Eq1) x == y => f x == f y, for all f of appropriate type which is analogous to this (existant) law about observational equality: (Eq2) x = y => f x = f y, for all f of appropriate type Kalman ---------------------------------------------------------------------- Finally - A spam blocker that actually works. http://www.bluebottle.com/tag/4