On 3/17/08, Hugo Pacheco <[EMAIL PROTECTED]> wrote: > On the other side, it fails to compile when this signature is explicit: > fff :: forall d x. (FunctorF d) => d -> F d x -> F d x > fff a = fmapF a id
Interestingly, this works when you also give a type signature to "id": fff :: forall d x. (FunctorF d) => d -> F d x -> F d x fff a = fmapF a (id :: x -> x) compiles for me (ghc6.8.2). There's probably a bug in the type checker; inference is working with no type signatures, but checking fails. _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
