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Corecursion 意味着在每次迭代中调用自己的数据大于或等于以前的数据。Corecursion 适用于 codata，它们是递归定义的值。不幸的是，值递归在严格评估的语言中是不可能的。不过，我们可以使用显式的 thunk：

const Defer = thunk =>
  ({get runDefer() {return thunk()}})

const app = f => x => f(x);

const fibs = app(x_ => y_ => {
  const go = x => y =>
    Defer(() =>
      [x, go(y) (x + y)]);

  return go(x_) (y_).runDefer;
}) (1) (1);

const take = n => codata => {
  const go = ([x, tx], acc, i) =>
    i === n
      ? acc
      : go(tx.runDefer, acc.concat(x), i + 1);

  return go(codata, [], 0);
};

console.log(
  take(10) (fibs));

虽然这按预期工作，但方法似乎很尴尬。尤其是可怕的对元组让我很恼火。有没有更自然的方法来处理 JS 中的 corecursion/codata？

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1 回答 1

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我会在数据构造函数本身中对 thunk 进行编码。例如，考虑。

// whnf :: Object -> Object
const whnf = obj => {
    for (const [key, val] of Object.entries(obj)) {
        if (typeof val === "function" && val.length === 0) {
            Object.defineProperty(obj, key, {
                get: () => Object.defineProperty(obj, key, {
                    value: val()
                })[key]
            });
        }
    }

    return obj;
};

// empty :: List a
const empty = null;

// cons :: (a, List a) -> List a
const cons = (head, tail) => whnf({ head, tail });

// fibs :: List Int
const fibs = cons(0, cons(1, () => next(fibs, fibs.tail)));

// next :: (List Int, List Int) -> List Int
const next = (xs, ys) => cons(xs.head + ys.head, () => next(xs.tail, ys.tail));

// take :: (Int, List a) -> List a
const take = (n, xs) => n === 0 ? empty : cons(xs.head, () => take(n - 1, xs.tail));

// toArray :: List a -> [a]
const toArray = xs => xs === empty ? [] : [ xs.head, ...toArray(xs.tail) ];

// [0,1,1,2,3,5,8,13,21,34]
console.log(toArray(take(10, fibs)));

这样，我们可以将惰性编码为弱头范式。优点是消费者不知道给定数据结构的特定字段是惰性的还是严格的，不需要关心。

于 2020-03-06T02:35:30.987 回答