More concretely, it's likely that Ord Bool is defined via `compare`, which is necessarily strict in both arguments.
Yes, this did come to mind. In general a non-strict `compare` would only make sense for `()`.
One of the relationships implied by the Ord typeclass is:
a <= b = True iff compare a b = EQ \/ compare a b = LT
So wouldn’t it make sense to define `compare` in terms of the “weaker” relations? It seems very unhaskelly to do the unnecessary work of evaluating the second argument to a relation when we already know what the result should be.
Vilem
On 2 Jan 2019, at 10:29, Isaac Elliott mailto:isaace71295@gmail.com> wrote:
One of the relationships implied by the Ord typeclass is:
a <= b = True iff compare a b = EQ \/ compare a b = LT
If we write an alternative definition of (<=) that is only strict in its first argument:
False <= _ = True
True <= x = x
Then it's impossible to write `compare` in a way that's consistent with that relation.
More concretely, it's likely that Ord Bool is defined via `compare`, which is necessarily strict in both arguments.
On Wed, 2 Jan. 2019, 7:47 pm V.Liepelt, mailto:V.Liepelt@kent.ac.uk> wrote:
I am surprised to find that `False <= undefined = undefined`.
What justifies (<=) to be strict in both arguments?
Vilem
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