>> f n () = (n, f (n + 1)) > In your example, if we assume that "f" has type, say, "a->()->(a,b)", for > some "a","b", then it is used, in its own definition, with type > "a->()->b".
Oops, should be: ... then it is used ... with type "a->b" (or Num a=>a->b). > ... "a->()->b" is not (cannot be) an instance of "a->()->(a,b)" ... Should be: "a->b" cannot be an instance of "a->()->(a,b)". _______________________________________________ Haskell-Cafe mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell-cafe
