[Haskell-cafe] Newbie question on Haskell type

[John Meacham]

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

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John Meacham - ⑆repetae.net⑆john⑈