受clojure启发的换能器可以用HM型系统打字吗

Question

我在Javascript中有一个纯功能的转换器实现，它支持回路融合和短路。请注意，尽管我使用的是JS，但这并不是理解问题的必要条件。只有类型才重要。

// ((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

正如你所看到的，折叠可以工作，但不能打字检查。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，不能键入吗？我在统一过程中犯了错误吗？还是我的转换器类型(map/filtert)完全错误？

Answer 1

map和filter的类型签名不够通用。

// 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));

acc的类型和k的输入类型应该相同，并且它们应该独立于其他类型。k的返回类型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只是转换为延续传递样式的cons的类型。使用此类型时，生成的程序类型将按预期进行检查。

// 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);

然而，将减速器转换为CPS并没有任何意义。将减速器转换为CPS不会给你带来任何好处，因为换能器无论如何都是用CPS编写的。事实上，在上面的程序中，CPS毫无意义，因为您使用的唯一延续是id(在foldk函数中)。如果你不使用CPS，那么你的程序就会变得简单得多。

// 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continuation而使用continuation。99%的时间你不需要继续。