[Haskell] Re: Rank-N types vs existential types

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/>