14 Oct
2005
14 Oct
'05
7:42 a.m.
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⑈