Because you could have an instance: instance RT r t1
and another instance for a t2 /= t1: instance RT r t2 then when you say: g (Dr r) = Dt (r t) then, 'r t' could either have type t1 or t2, thus giving the result value type 'Dt t1' or 'Dt t2'. If your class had fundeps 'r -> t', then this probably wouldn't be necessary. -- Hal Daume III "Computer science is no more about computers | [EMAIL PROTECTED] than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume On Mon, 27 Jan 2003, Dean Herington wrote: > Can someone explain why the type declaration for `g` is required in the > following? > > > class RT r t where rt :: r -> t > > data D t = Dt t | forall r. RT r t => Dr r > > f :: D t -> D t > f = g > where -- g :: D t -> D t > g (Dr r) = Dt (rt r) > > > As given above, the program evokes these error messages: > > with GHC 5.04.2: > Bug4.hs:10: > Could not deduce (RT r t1) from the context (RT r t) > Probable fix: > Add (RT r t1) to the existential context of a data constructor > arising from use of `rt' at Bug4.hs:10 > In the first argument of `Dt', namely `(rt r)' > In the definition of `g': Dt (rt r) > > with Hugs: > ERROR "Bug4.hs":6 - Cannot justify constraints in explicitly typed > binding > *** Expression : f > *** Type : D a -> D a > *** Given context : () > *** Constraints : RT _6 a > > > -- Dean > > _______________________________________________ > Haskell mailing list > [EMAIL PROTECTED] > http://www.haskell.org/mailman/listinfo/haskell > _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
