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