Subject: |
Re: [Haskell-cafe] Re: (flawed?) benchmark : sort |
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From: |
Dan Weston |

Date: |
Mon, 10 Mar 2008 14:01:53 -0700 |

If x <= y && y <= x does not imply that x == y, then Ord has no business
being a subclass of Eq. By your logic, there is absolutely no
constructive subclassing going on here, only an existence proof of (==)
given (<=). What is the rational basis of such an existence claim,
unless == has the obvious meaning?
Or should I take it that you are suggesting we should move Ord up to be
a peer of Eq?
Dan Neil Mitchell wrote: HiBut antisymmetry means that (x <= y) && (y <= x) ==> x = y, where '=' means identity. Now what does (should) 'identity' mean?I think you are using the word identity when the right would would be equality. Hence, the answer is, without a doubt, (==). If you define equality, then you are defining equality.In short, I would consider code where for some x, y and a function f we have (x <= y) && (y <= x) [or, equivalently, compare x y == EQ] but f x /= f y broken indeed.I would consider it slightly bad code too. But not broken code. I think Ord functions can assume that Ord is a total ordering, nothing more. Nothing to do with the existence (or otherwise) of entirely unrelated code. Consider the following implementation of Data.Set, which *does* meet all the invariants in Data.Set: data Set a = Set [a] insert x (Set xs) = Set $ x : filter (/= x) xs elems (Set xs) = xs i.e. there is real code in the base libraries which breaks this notion of respecting classes etc. Is the interface to Data.Set broken? I would say it isn't. Thanks Neil _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe |

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