
15 Jul
2010
15 Jul
'10
7:09 a.m.
On Thu, Jul 15, 2010 at 3:13 AM, Michael Snoyman
You could also use type families for this, but I believe you cannot express the "Show" superclass: class MyClass a where type MyResult a fn :: a -> MyResult a
I guess this works: class Show (MyResult a) => MyClass a where type MyResult a fn :: a -> MyResult a HTH, -- Felipe.