At Tue, 15 Jul 2008 10:27:59 -0400, Jeff Polakow wrote:

[1 ]

[2 ] Hello,

...

data LSet t where Nil :: LSet Nil Ins :: (Member a t b , If b t (a ::: t) r) => L a -> LSet t -> LSet r

Try replacing both original occurrences of r, i.e. (untested)

Ins :: (Member a t b, If b t (a ::: t) (LSet r)) => L a -> LSet t -> LSet r

Thanks. This sort of works, but shifts the problem to another context. Now it seems that I can't hide the extra type information in the existential types, which is what I want to do. In the function `insertChar' below I want the type LSetBox to be opaque (i.e. it will be called by users who don't need to know about the fancy types): data LSet t where Nil :: LSet Nil Ins :: (Member a t b , If b t (a ::: t) (LSet r)) => L a -> LSet t -> LSet r -- I have to supply a type for `insert' now and it must include the constraints insert :: (Member a t b , If b t (a ::: t) (LSet r)) => L a -> LSet t -> LSet r insert = Ins --insertChar (and the boxing) doesn't work insertChar :: Char -> LSetBox -> LSetBox insertChar c (LSetBox s) = case fromChar c of LBox t -> LSetBox (insert t s) The error: Could not deduce (If b t1 (a ::: t1) (LSet t), Member a t1 b) from the context () arising from a use of `insert' at /home/jim/sdf-bzr/dsel/TF/Set-July08.hs:54:25-34 Possible fix: add (If b t1 (a ::: t1) (LSet t), Member a t1 b) to the context of the constructor `LBox' or add an instance declaration for (If b t1 (a ::: t1) (LSet t)) In the first argument of `LSetBox', namely `(insert t s)' In the expression: LSetBox (insert t s) In a case alternative: LBox t -> LSetBox (insert t s) Failed, modules loaded: none. Jim

-Jeff

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