I think I get it. But to be sure: The forall means: The type for a may not be the same throughout the whole function. It just has to follow the pattern specified inside the braces.
Interestingly, when I want hugs to show me the type of > fun::(forall a.[a]->Int)->[b]->[c]->Int it tells me: ERROR - Use of fun requires at least 1 argument Why that? At least I have explicitely specified the type. Markus > One may want: > > fun f x y = f x + f y > > for instance: > > fun length [True, False] [1,2] > > Therefore, I would say, you need typing a la > > fun::(forall a.[a]->Int)->[b]->[c]->Int > > Ciao, > Steffen > _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
