On Fri, Oct 14, 2005 at 03:17:12AM -0400, Cale Gibbard wrote:
> Right, forgot about seq there, but the point still holds that there
> are a very limited number of functions of that type, and in
> particular, the functions can't decide what to do based on the type
> parameter 'a'.
>
actually, without 'seq' _|_ and \_ -> _|_ are indistinguishable. so you
only have 3 functions without seq, and 6 with it.
John
--
John Meacham - ⑆repetae.net⑆john⑈
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe
