您如何在Clojure的core.typed中注释多态核心功能？

Question

我想在代码中应用core.type批注，但是遇到了如何/何时实例化多态的核心函数（从函数主体内部调用）的绊脚石。

通过对此进行故障排除，我了解到必须对filter和count进行特殊处理，因为它们分别是多态的和静态的，应在let绑定中提取匿名函数并对其进行注释。 如果有人可以根据以下错误消息的输出解释如何对此进行注释，我将非常感激。

这是我的别名：

(defalias Key-set (Set (Option Kw)))
(defalias Player (Ref1 Key-set))
(defalias Set-vec (Vec (Set Kw)))
(defalias Move (U ':x ':o))

该函数的当前形式：

    (ann first-two [Player -> (Option Seq)])
    (defn first-two [player]
      (let [;; best guesses here
            filter>    (inst filter [[(U (Seqable Any) clojure.lang.Counted nil) -> Bool] -> Any] (Option (clojure.lang.Seqable [(U (Seqable Any) clojure.lang.Counted nil) -> Bool])))
            count>     (ann-form count [(U nil (Seqable Any) clojure.lang.Counted) -> Number])

            two-count? :- [(U nil (Seqable Any) clojure.lang.Counted)-> Bool], #(= (count> %) 2)
            couples    :- (Option (Seqable Key-set)), (concat adjacents opposite-corners opposite-sides)]
        (first (filter> two-count?
                        (for [pair :- Key-set, couples]
                          (clojure.set/intersection pair @player))))))

类型检查器错误消息：

Type Error (tic_tac_toe/check.clj:52:5) Function filter> could not be applied to arguments:


Domains:
    [[[(U (Seqable Any) Counted nil) -> Bool] -> Any] -> Any :filters {:then (is (Option (clojure.lang.Seqable [(U nil (Seqable Any) clojure.lang.Counted) -> Bool])) 0), :else tt}] (Option (clojure.lang.Seqable [[(U (Seqable Any) Counted nil) -> Bool] -> Any]))
    [[[(U (Seqable Any) Counted nil) -> Bool] -> Any] -> Any :filters {:then (! (Option (clojure.lang.Seqable [(U nil (Seqable Any) clojure.lang.Counted) -> Bool])) 0), :else tt}] (Option (clojure.lang.Seqable [[(U (Seqable Any) Counted nil) -> Bool] -> Any]))
    [[[(U (Seqable Any) Counted nil) -> Bool] -> Any] -> Any] (Option (clojure.lang.Seqable [[(U (Seqable Any) Counted nil) -> Bool] -> Any]))

Arguments:
    [(U nil (Seqable Any) clojure.lang.Counted) -> Bool] (clojure.lang.LazySeq Any)

Ranges:
    (ASeq (Option (clojure.lang.Seqable [(U nil (Seqable Any) clojure.lang.Counted) -> Bool])))
    (ASeq (I [[(U (Seqable Any) Counted nil) -> Bool] -> Any] (Not (Option (clojure.lang.Seqable [(U nil (Seqable Any) clojure.lang.Counted) -> Bool])))))
    (ASeq [[(U (Seqable Any) Counted nil) -> Bool] -> Any])

ExceptionInfo Type Checker: Found 1 error  clojure.core/ex-info (core.clj:4566)

原始功能：

(defn first-two [player]
  (let [couples (concat adjacents opposite-corners opposite-sides)]
    (first (filter #(= (count %) 2)
                   (for [pair couples]
                     (clojure.set/intersection pair @player))))))

Answer 1

您可以在其他地方提供类型以避免实例化filter 。

例如，此类型检查。

(ns typed-test.core
  (:refer-clojure :exclude [fn for])
  (:require [clojure.core.typed :as t
             :refer [fn for Vec Coll Any Num]]))

(filter (fn [a :- (Coll Any)] (= (count a) 2))
        (for [pair :- (Vec Num), {1 2 3 4}] 
          :- (Vec Num)
          pair))

也许缺少调用的返回类型for是失去了太多的信息。

inst只是用某些类型替换All绑定程序中的类型变量。

typed-test.core=> (cf identity)
(t/All [x] [x -> x :filters {:then (! (t/U nil false) 0), :else (is (t/U nil false) 0)} :object {:id 0}])
typed-test.core=> (cf (t/inst identity Num))
[Num -> Num :filters {:then (! (t/U nil false) 0), :else (is (t/U nil false) 0)} :object {:id 0}]

Answer 2

只是一个简单的主意，因为我现在没有时间去找你。

但是，通常在类型化Clojure中的多态类型参数不会直接映射到参数。 快速浏览一下，我认为这就是问题的根源。

例如，假设我们定义了一个多态的fun并且想要实例化它。

(ann f (All [x] [[x -> Bool] x -> Integer]))
(defn f [predicate? value] (if (f value) 1 0))

要正确实例化它，您必须像那样

((inst f String) (typed/fn [x :- String] :- Bool true) "lol")

并不是

((inst f [String -> Bool]) (typed/fn [x :- String] :- Bool true) "lol2") ;; baaaad baaaad

现在我必须回去工作，今天花了太多时间定制我的emacs啊哈...