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1. Previously, the following term wouldn't type-check: match x = term { true: ct false: cf } When `term` isn't a variable. The reason is that the `match` syntax was desugared to: (~term _ ct cf) Which, in general, results in the following check: true : ∀(P: Bool -> *) ∀(t: P true) ∀(f: P false) (P term) check _ :: Bool -> * check ct :: (_ true) check cf :: (_ false) check () :: (_ term) unify (_ term) == (P term) This should generate the solution: _ = λx(P x) But, when `term` isn't a variable (it could be a function call, or a ref, for example), this fails to satisfy the pattern unification criteria. To fix that, we just use a `let` to always create a variable before unifying, so that the `match` syntax is desugared to: let x = term (~x _ ct cf) Which solves the issue. 2. Previously, the following term wouldn't type-check: equal (a: Nat) (b: Nat) : Bool = match a with (b: Nat) { succ: match b { succ: (equal a.pred b.pred) zero: false } zero: match b { succ: false zero: true } } After inspecting the type-checker output, I noticed the reason is due to how the outer hole interacts with the inner hole. Specifically, after some computations we eventually get to the following unification problem: unify _2 == (_3 b) Here, there are two ways to solve this: _2 = (_3 b) or _3 = λx (_2) Initially, it seems like both solutions are equivalent. Yet, by choosing the first, we, later on, face the following problem: - (_3 (succ x)) - Bool Which violates the pattern unification requisites (for the same reason: `succ` is a constructor, rather than a variable). A way to solve this problem would be to simply invert the order, and use `_3 = λx (_2)` instead (done in the last commit). That way, we'd have `(_3 (succ x))` reduce to just `_2`, which would solve `_2 = Bool` and type-check fine. This is a quickfix though. I wonder if there is a more principled approach that would let us avoid `(succ x)` to be there at all. Debug log: https://gist.github.com/VictorTaelin/6dd1a7167d5b7c8d5626c0f5b88c75a0
8 lines
135 B
Plaintext
8 lines
135 B
Plaintext
use Bool/{true,false}
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abs_diff (a: U60) (b: U60) : U60 =
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match x = (U60/to_bool (< a b)) {
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true: (- b a)
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false: (- a b)
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}
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