Data.Type.Equality.== works better when used at kind * -> * -> Bool
Richard Eisenberg
eir at cis.upenn.edu
Sat May 31 21:21:19 UTC 2014
On May 31, 2014, at 1:12 AM, adam vogt <vogt.adam at gmail.com> wrote:
> are
> there instances of (==) that behave differently from the poly-kinded
> version?
Yes.
To be concrete, here would be the polykinded instance:
> type family EqPoly (a :: k) (b :: k) where
> EqPoly a a = True
> EqPoly a b = False
> type instance (a :: k) == (b :: k) = EqPoly a b
Note that this overlaps with every other instance -- if this were defined, it would be the only instance for (==).
Now, consider
> data Nat = Zero | Succ Nat
Suppose I want
> foo :: (Succ n == Succ m) ~ True => ((n == m) :~: True)
> foo = Refl
This would not type-check with the poly-kinded instance. `Succ n == Succ m` quickly becomes `EqPoly (Succ n) (Succ m)` but then is stuck. We don't know enough about `n` and `m` to reduce further.
On the other hand, consider this:
> type family EqNat (a :: Nat) (b :: Nat) where
> EqNat Zero Zero = True
> EqNat (Succ n) (Succ m) = EqNat n m
> EqNat n m = False
> type instance (a :: Nat) == (b :: Nat) = EqNat a b
With this instance, `foo` type-checks fine. `Succ n == Succ m` becomes `EqNat (Succ n) (Succ m)` which becomes `EqNat n m`. Thus, we can conclude `(n == m) ~ True` as desired.
So, the Nat-specific instance allows strictly more reductions, and is thus preferable to the poly-kinded instance. But, if we introduce the poly-kinded instance, we are barred from writing the Nat-specific instance, due to overlap.
Even better than the current instance for * would be one that does this sort of recursion for all datatypes, something like this:
> type family EqStar (a :: *) (b :: *) where
> EqStar Bool Bool = True
> EqStar (a,b) (c,d) = a == c && b == d
> EqStar (Maybe a) (Maybe b) = a == b
> ...
> EqStar a b = False
The problem is the (...) is extensible -- we would want to add new cases for all datatypes in scope. This is not currently possible for closed type families. Perhaps it would be an improvement to write the cases for all types in scope in Data.Type.Equality -- that is, the types exported from `base`.
I hope this is helpful. In any case, I will put some of the text in this email into the comments in Data.Type.Equality for the next person who looks.
Richard
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