[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
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.
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