多态变体上的通配符模式覆盖子类型约束

问题描述

给出这些类型

type a = [ `A ]
type b = [ a | `B  | `C ]

和此功能

let pp: [< b] -> string =
  function | `A -> "A"
           | `B -> "B"
           | `C -> "C"

应用类型为a的值可以正常工作，如预期:

applying a value of type a works without issue, as expected:

let a: a = `A
let _ = pp a

但是，如果将功能修改为包含通配符模式

However, if the function is modified to include a wildcard pattern

let pp: [< b] -> string =
  function | `A -> "A"
           | `B -> "B"
           | _ -> "?"

即使其他所有内容保持不变，它现在也会产生以下错误(在let _ = pp a上):

even though everything else remains the same, it now yields the following error (on let _ = pp a):

问题:

1. 为什么它不再能够接受子类型?我知道通配符意味着它现在可以接受超类型，但这并不意味着必须.
2. 是否有某种方法可以避免这种情况，而不必枚举一百万个不相关的变体?

推荐答案

基本问题是为什么

let pp= function
| `A -> "A"
| `B -> "B"
| _ -> "?"

推断为[> `A| `B] -> string而不是[< `A| `B | ... ] -> string(其中...代表任何构造函数).答案是设计选择和在误报与误报之间折衷的问题: https://www.math.nagoya-u.ac.jp/~garrigue/papers/matching.pdf .

is infered as [> `A| `B] -> string and not as [< `A| `B | ... ] -> string (where ... stands for any constructor). The answer is that is a design choice and a question of compromise between false positive and false negative : https://www.math.nagoya-u.ac.jp/~garrigue/papers/matching.pdf .

更准确地说，第二种类型被认为太弱了，因为它很容易丢失pp中存在`A和`B的信息.例如，考虑以下代码，其中`b是拼写错误，应该是`B:

More precisely, the second type was deemed too weak since it was too easy to lose the information that `A and `B were present in pp. For instance, consider the following code where `b is a spelling mistake and should have been `B:

let restrict (`A | `b) = ()
let dual x = restrict x, pp x

当前，此代码失败

这时，如果`b是拼写错误，则可以在此处捕获该错误.如果pp被键入[< `A|`B |..]，则对偶类型将被无声地限制为[`A] -> unit * string，没有机会捕获此错误.而且，在当前键入的情况下，如果`b不是拼写错误，则完全有可能通过添加一些强制性来使dual有效

At this point, if `b was a spelling mistake, it becomes possible to catch the mistake here. If pp had been typed [< `A|`B |..], the type of dual would have been restricted to [`A] -> unit * string silently, with no chance of catching this mistake. Moreover, with the current typing, if `b was not a spelling mistake, it is perfectly possible to make dual valid by adding some coercions

let dual x = restrict x, pp (x:[`A]:>[>`A]);;
(* or *)
let dual x = restrict x, (pp:>[`A] -> _) x

非常明确地说明restrict和pp可在不同的多态变体集上工作.

making it very explicit that restrict and pp works on different sets of polymorphic variants.

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