hello, just skip the forall. "free" type variables are implicitly forall-quantified. the forall keyword is used for functions that need to take polymorphic functions as arguments, but that's advanced stuff. so in short, this should work:

class Space t class Space t => HasY v t where assignY :: v -> t -> t getY :: t -> v varY :: Space w => v -> (t->t) -> (w->w)

the next problem you are likely to run into is ambiguities, and you might want to take a look at functional dependencies. hope this helps iavor Theodore S. Norvell wrote:

Can anyone explain the following error message from Hug98 (with extensions enabled)?

I have

...

class Space t class Space t => HasY v t where assignY :: v -> t -> t getY :: t -> v varY :: forall w . Space w => v -> (t->t) -> (w->w)

I get an error on the last definition "quantifier does not mention type variable v" (and if I quantify v, then it complains that t is not quantified).

The idea I'm trying to express is that for any instance of HasY v t there should be a function varY with type v -> (t->t) -> (w->w) for some type w of class Space.

Any help much appreciated.

Is there any kind of tutorial introduction to "forall" out there?

Cheers, Theo Norvell

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-- ================================================== | Iavor S. Diatchki, Ph.D. student | | Department of Computer Science and Engineering | | School of OGI at OHSU | | http://www.cse.ogi.edu/~diatchki | ==================================================