[Haskell] Re: Rank-N types vs existential types
ozone at algorithm.com.au
Wed Apr 27 07:56:00 EDT 2005
On 27/04/2005, at 9:35 PM, Andreas Rossberg wrote:
> The difference really shows up when you *use* (deconstruct) them:
> g (RankNEq f) = (f 4 5, f True False)
> This allows the embedded function to be used polymorphically. But:
> h (ExistsEq f) = ???
> Here, you cannot use f at all (well, except with undefined). The type
> is not polymorphic in "a" on the RHS, it is abstract! You'd need to
> encapsulate a value of the same type (or a constructing function) as
> well to this type useful.
Ah, OK! That's quite enlightening. Thanks Andreas, Krasimir, and
Tomasz for all your replies.
% Andre Pang : trust.in.love.to.save <http://www.algorithm.com.au/>
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