[Haskell-cafe] A question about GHC test coverage

Olaf Klinke olf at aatal-apotheke.de
Tue Feb 21 20:25:20 UTC 2017

> Am 20.02.2017 um 23:05 schrieb Richard A. O'Keefe <ok at cs.otago.ac.nz>:
> On 15/02/17 10:35 AM, Olaf Klinke wrote:
>> You should not have to write tests for functions you did not define.
> Correct me if I'm wrong, but any property of max can be derived from the properties of <=.
> Unfortunately, this isn't quite true.
> Suppose we have two values x y such that x <= y && y <= x && x == y
> *BUT* x and y are distinguishable some other way.  For example,
> suppose we are modelling rational numbers by pairs (n,d) *without*
> insisting that gcd(n,d) == 0.  Then we have
>   (n1,d1) == (n2,d2) = n1*d2 == n2*d1
>   compare (n1,d1) (n2,d2) = compare (n1*d2) (n2*d1)
> BUT (1,2) and (2,4), while ==, are none-the-less distinguishable.
> Now ask about max x y.  If we have
>    max a b = if a > b then a else b
> then max x y delivers y.  But if we have
>    max a b = if a < b then b else a
> then max x y delivers x, and these two results can be distinguished.
If you define an Ord instance that is only a preorder rather than a total order, and then downstream rely on functions f that violate the property

x == y implies f x == f y,

I'd call that poor design. The above formula would be something for a Quickcheck module. That said, I myself found it handy in the past to define such preordered types.
> I know and freely admit that if I want a version of max that satisfies
> stronger conditions than Ord can guarantee, it is up to me to write it.
Or rather, be aware of the standard implementation of max in Data.Ord. 
Anyways, you are right: Since the type system can not guarantee that every Ord instance is actually a total order, test coverage statistics won't help us here. 


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