[Haskell-cafe] Puzzled by error with forall quantifier.

Theodore S. Norvell theo at engr.mun.ca
Mon Feb 2 16:55:08 EST 2004

Iavor S. Diatchki wrote:

> just skip the forall.   "free" type variables are implicitly 
> forall-quantified.

Of course!  Thanks.  The problem is that I actually wanted "exists".
However as the type that exists is functionally dependent on other the
parameters to the class I think I can deal with it using functional
dependencies, as you suggest.

Sorry for the elementary question.

Theo Norvell

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