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我有一个纯功能的 Javascript 转换器实现，支持循环融合和短路。请注意，虽然我使用的是 JS，但这并不是理解问题的必要条件。只有类型很重要。

// ((a -> r) -> r) -> Cont r a
const Cont = k => ({run: k});

// (a -> b -> Cont b b) -> b -> [a] -> b
const foldk = f => init => xs => {
  let acc = init;

  for (let i = xs.length - 1; i >= 0; i--)
    acc = f(xs[i]) (acc).run(id);

  return acc;
};

// (a -> b) -> (b -> t b -> Cont (t b) (t b)) -> a -> t b -> Cont (t b) (t b)
const map = f => cons => x => acc =>
  Cont(k => cons(f(x)) (acc).run(k));

// (a -> Bool) -> (a -> t a -> Cont (t a) (t a)) -> a -> t a -> Cont (t a) (t a)
const filter = p => cons => x => acc =>
  Cont(k =>
    p(x)
      ? cons(x) (acc).run(k)
      : k(acc));

// (b -> c) -> (a -> b) -> a -> c
const comp = f => g => x => f(g(x));

const liftCont2 = f => x => y => Cont(k => k(f(x) (y)));
const unshift = x => xs => (xs.unshift(x), xs);
const includes = sub => s => s.includes(sub);
const len = arrayLike => arrayLike.length;
const id = x => x;

// (String -> t String -> Cont (t String) (t String))
//   -> String -> t String -> Cont (t String) (t String)
const filterO = filter(includes("o"));

// (Number -> t Number -> Cont (t Number) (t Number))
//    -> String -> t Number -> Cont (t Number) (t Number)
const mapLen = map(len);

// type error
const transducer = comp(filterO) (mapLen);

const reducer =  transducer(liftCont2(unshift));

const main = foldk(reducer) ([]) (["f", "fo", "foo", "fooo"]);

console.log(main); // [2, 3, 4] with only a single traversal

如您所见，折叠有效，但不进行类型检查。揭示类型错误需要一些手动统一：

unify `comp(filterO)`:
(b -> c) -> (a -> b) -> a -> c
(String -> t String -> Cont (t String) (t String))
  -> String -> t String -> Cont (t String) (t String)

yields

(a -> String -> t String -> Cont (t String) (t String))
  -> a -> String -> t<String> -> Cont (t String) (t String)

unify result of `comp(filterO)` with `mapLen` (contravariant):
a -> String -> t String -> Cont (t String) (t String)
(Number -> t Number -> Cont (t Number) (t Number))
  -> String -> t Number -> Cont (t Number) (t Number)

substitutes

a ~ Number -> t Number -> Cont (t Number) (t Number)

unify (covariant):
String -> t String -> Cont (t String) (t String)
String -> t Number -> Cont (t Number) (t Number)

这两个术语显然不能统一。

转换器是动态语言独有的概念并且无法键入，是我在统一过程中犯了错误还是我的转换器类型（map/ filtert）完全错误？

4

1 回答 1

2

你的类型签名map和filter不够通用。

// map :: (a -> b) -> (b -> t b -> Cont (t b) (t b)) -> a -> t b -> Cont (t b) (t b)
const map = f => cons => x => acc => Cont(k => cons(f(x))(acc).run(k));

// filter :: (a -> Bool) -> (a -> t a -> Cont (t a) (t a)) -> a -> t a -> Cont (t a) (t a)
const filter = p => cons => x => acc => Cont(k => p(x) ? cons(x)(acc).run(k) : k(acc));

的类型acc和输入的类型k应该相同，并且它们应该独立于其他类型。的返回类型k也应该独立于其他类型。

// type Reducer r a b = a -> b -> Cont r b

// map :: (a -> b) -> Reducer r b c -> Reducer r a c
const map = f => cons => x => acc => Cont(k => cons(f(x))(acc).run(k));

// filter :: (a -> Bool) -> Reducer r a b -> Reducer r a b
const filter = p => cons => x => acc => Cont(k => p(x) ? cons(x)(acc).run(k) : k(acc));

请注意，类型Reducer只是cons转换为延续传递样式的类型。使用此类型时，生成的程序类型会按预期进行检查。

// data Cont r a = Cont { run :: (a -> r) -> r }
const Cont = run => ({ run });

// type Reducer r a b = a -> b -> Cont r b

// foldk :: Reducer b a b -> b -> [a] -> b
const foldk = f => init => xs => xs.reduceRight((acc, x) => f(x)(acc).run(id), init);

// map :: (a -> b) -> Reducer r b c -> Reducer r a c
const map = f => cons => x => acc => Cont(k => cons(f(x))(acc).run(k));

// filter :: (a -> Bool) -> Reducer r a b -> Reducer r a b
const filter = p => cons => x => acc => Cont(k => p(x) ? cons(x)(acc).run(k) : k(acc));

// comp :: (b -> c) -> (a -> b) -> a -> c
const comp = f => g => x => f(g(x));

// liftCont2 :: (a -> b -> c) -> a -> b -> Cont r c
const liftCont2 = f => x => y => Cont(k => k(f(x)(y)));

// unshift :: a -> [a] -> [a]
const unshift = x => xs => [x, ...xs];

// includes :: String -> String -> Bool
const includes = sub => s => s.includes(sub);

// len :: ArrayLike t => t a -> Number
const len = arrayLike => arrayLike.length;

// id :: a -> a
const id = x => x;

// filterO :: Reducer r String a -> Reducer r String a
const filterO = filter(includes("o"));

// mapLen :: ArrayLike t => Reducer r Number b -> Reducer r (t a) b
const mapLen = map(len);

// transducer :: Reducer r Number a -> Reducer r String a
const transducer = comp(filterO)(mapLen);

// reducer :: Reducer r String [Number]
const reducer = transducer(liftCont2(unshift));

// main :: [Number]
const main = foldk(reducer)([])(["f", "fo", "foo", "fooo"]);

// [2, 3, 4]
console.log(main);

但是，将reducer转换为CPS并没有任何意义。将减速器转换为 CPS 不会给您带来任何好处，因为传感器无论如何都是用 CPS 编写的。事实上，在上述程序中，CPS 毫无意义，因为您使用的唯一延续是id（在foldk函数中）。如果您不使用 CPS，那么您的程序会变得简单得多。

// type Reducer a b = a -> b -> b

// foldr :: Reducer a b -> b -> [a] -> b
const foldr = f => init => xs => xs.reduceRight((acc, x) => f(x)(acc), init);

// map :: (a -> b) -> Reducer b c -> Reducer a c
const map = f => cons => x => cons(f(x));

// filter :: (a -> Bool) -> Reducer a b -> Reducer a b
const filter = p => cons => x => p(x) ? cons(x) : id;

// comp :: (b -> c) -> (a -> b) -> a -> c
const comp = f => g => x => f(g(x));

// unshift :: a -> [a] -> [a]
const unshift = x => xs => [x, ...xs];

// includes :: String -> String -> Bool
const includes = sub => s => s.includes(sub);

// len :: ArrayLike t => t a -> Number
const len = arrayLike => arrayLike.length;

// id :: a -> a
const id = x => x;

// filterO :: Reducer String a -> Reducer String a
const filterO = filter(includes("o"));

// mapLen :: ArrayLike t => Reducer Number b -> Reducer (t a) b
const mapLen = map(len);

// transducer :: Reducer Number a -> Reducer String a
const transducer = comp(filterO)(mapLen);

// reducer :: Reducer String [Number]
const reducer = transducer(unshift);

// main :: [Number]
const main = foldr(reducer)([])(["f", "fo", "foo", "fooo"]);

// [2, 3, 4]
console.log(main);

不要仅仅为了使用延续而使用延续。99% 的时间你不需要继续。

于 2021-08-03T08:52:39.960 回答