On Mon, Jul 15, 2013 at 11:21:39PM -0400, Clark Gaebel wrote:
As a tangent, can anyone think of a data structure for which you can write an Ord instance but Hashable/Eq is impossible (or prove otherwise)? How about the converse?
A conversation on #haskell brought back to mind this question, which I'm not sure was ever answered. Is there a type with an Eq instance for which on Ord instance could not also be provided (i.e. I'm interested in Clark's "converse")? One answer is that I could write data A = A deriving Eq and then not export the constructor. That would make it impossible to write an Ord instance. However, I'm more interested in the question "morally" or "in principle". If we think about algebraic datatypes built from sums and products of base types, which have both Eq and Ord, and the function type construct, which prevents both Eq and Ord, it seems that the two always go together, in principle. Am I missing anything more exotic? IORef's have Eq but not Ord. Is there a good reason for that? Would Ord break "referential transparency" somehow? I doubt it. How about STRef? It seems more likely they shouldn't have Ord. Any thoughts? Tom