[Haskell-cafe] Re: (flawed?) benchmark : sort

[Daniel Fischer]

Am Montag, 10. März 2008 21:34 schrieb Neil Mitchell:
> Hi
>
> >  But 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.

Okay, bad choice of words. Of course I expect 
compare x y == EQ <==> x == y 
for any Ord instance.
And for
f :: (Eq a, Eq b) => a -> b
I expect (x == y) ==> (f x == f y).
>
> >  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

Perhaps 'broken' is a stronger word than I thought. I wouldn't say there can 
never be a reason for such. It would not necessarily be *badly* broken, 
though at the moment I can't see a case where such behaviour (in an exported 
function) would be reasonable. Of course, internal fuctions are a different 
matter.

> 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.

I would say, if we have x = Set [1,2], y = Set [2,1] and an Eq instance where
x == y, then elems shouldn't be exported.
>
> Thanks
>
> Neil

Cheers,
Daniel