如何折叠非尾递归算法的数据结构

Question

我有一个可变参数提升函数，它允许没有深度嵌套函数组合的扁平一元链：

const varArgs = f => {
  const go = args =>
    Object.defineProperties(
      arg => go(args.concat(arg)), {
        "runVarArgs": {get: function() {return f(args)}, enumerable: true},
        [TYPE]: {value: "VarArgs", enumerable: true}
      });
  return go([]);
};
const varLiftM = (chain, of) => f => { // TODO: replace recursion with a fold
  const go = (ms, g, i) =>
    i === ms.length
      ? of(g)
      : chain(ms[i]) (x => go(ms, g(x), i + 1));
  return varArgs(ms => go(ms, f, 0));
};

它有效，但我想用折叠从递归中抽象出来。正常的折叠似乎不起作用，至少与Task类型一起不起作用，

const varLiftM = (chain, of) => f =>
  varArgs(ms => of(arrFold(g => mx => chain(mx) (g)) (f) (ms))); // A

因为第 A 行中的代数将为每个迭代返回一个Task，而不是部分应用的函数。

如何将非尾递归替换为折叠？

下面是当前递归实现的工作示例：

const TYPE = Symbol.toStringTag;
const struct = type => cons => {
  const f = x => ({
    ["run" + type]: x,
    [TYPE]: type,
  });
  return cons(f);
};
// variadic argument transformer
const varArgs = f => {
  const go = args =>
    Object.defineProperties(
      arg => go(args.concat(arg)), {
        "runVarArgs": {get: function() {return f(args)}, enumerable: true},
        [TYPE]: {value: "VarArgs", enumerable: true}
      });
  return go([]);
};
// variadic monadic lifting function
const varLiftM = (chain, of) => f => { // TODO: replace recursion with a fold
  const go = (ms, g, i) =>
    i === ms.length
      ? of(g)
      : chain(ms[i]) (x => go(ms, g(x), i + 1));
  return varArgs(ms => go(ms, f, 0));
};
// asynchronous Task
const Task = struct("Task") (Task => k => Task((res, rej) => k(res, rej)));
const tOf = x => Task((res, rej) => res(x));
const tMap = f => tx =>
  Task((res, rej) => tx.runTask(x => res(f(x)), rej));
const tChain = mx => fm =>
  Task((res, rej) => mx.runTask(x => fm(x).runTask(res, rej), rej));
// mock function
const delay = (ms, x) =>
  Task(r => setTimeout(r, ms, x));
// test data
const tw = delay(100, 1),
  tx = delay(200, 2),
  ty = delay(300, 3),
  tz = delay(400, 4);
// specialization through partial application
const varAsyncSum =
  varLiftM(tChain, tOf) (w => x => y => z => w + x + y + z);
// MAIN
varAsyncSum(tw) (tx) (ty) (tz)
  .runVarArgs
  .runTask(console.log, console.error);
console.log("1 sec later...");

[编辑] 正如评论中所希望的那样，我的折叠实现：

const arrFold = alg => zero => xs => {
  let acc = zero;
  for (let i = 0; i < xs.length; i++)
    acc = alg(acc) (xs[i], i);
  return acc;
};

Answer 1

of arrFold周围打电话似乎有点不合适。

我不确定您的arrFold是右折叠还是左折叠，但假设它是右折叠，您将需要使用带有闭包的延续传递样式，就像您在递归实现中所做的那样：

varArgs(ms => of(arrFold(g => mx => chain(mx) (g)) (f) (ms)))

成为

varArgs(ms => arrFold(go => mx => g => chain(mx) (x => go(g(x)))) (of) (ms) (f))

---

用左折，你可以写

varArgs(arrFold(mg => mx => chain(g => map(g) (mx)) (mg)) (of(f)))

但你需要注意，这构建了一个与右折叠不同的调用树：

of(f)
chain(of(f))(g0 => map(m0)(g0))
chain(chain(of(f))(g0 => map(m0)(g0)))(g1 => map(m1)(g1))
chain(chain(chain(of(f))(g0 => map(m0)(g0)))(g1 => map(m1)(g1)))(g2 => map(m2)(g2))

vs(已应用续集(

of(f)
chain(m0)(x0 => of(f(x0)))
chain(m0)(x0 => chain(m1)(x1 => of(f(x0)(x1))))
chain(m0)(x0 => chain(m1)(x1 => chain(m2)(x2) => of(f(x0)(x1)(x2)))))

根据单子法，它们应该评估相同，但在实践中，一个可能比另一个更有效率。

Answer 2

对于这个特定的用例，你不需要monads的全部功能。应用函子是你所需要的：

// type Cont r a = (a -> r) -> r
// type Async a = Cont (IO ()) a
// pure :: a -> Async a
const pure = a => k => k(a);
// ap :: Async (a -> b) -> Async a -> Async b
const ap = asyncF => asyncA => k => asyncF(f => asyncA(a => k(f(a))));
// delay :: (Number, a) -> Async a
const delay = (ms, a) => k => setTimeout(k, ms, a);
// async1, async2, async3, async4 :: Async Number
const async1 = delay(100, 1);
const async2 = delay(200, 2);
const async3 = delay(300, 3);
const async4 = delay(400, 4);
// sum :: Number -> Number -> Number -> Number -> Number
const sum = a => b => c => d => a + b + c + d;
// uncurry :: (a -> b -> c) -> (a, b) -> c
const uncurry = f => (a, b) => f(a)(b);
// result :: Async Number
const result = [async1, async2, async3, async4].reduce(uncurry(ap), pure(sum));
// main :: IO ()
result(console.log);
console.log("1 second later...");

如果需要，您可以定义一个应用上下文函数(即 apply ( 如下：

const apply = (asyncF, ...asyncArgs) => asyncArgs.reduce(uncurry(ap), asyncF);
const result = apply(pure(sum), async1, async2, async3, async4);

如果你把这个函数，那么你可以创建一个lift函数：

const apply = asyncF => (...asyncArgs) => asyncArgs.reduce(uncurry(ap), asyncF);
const lift = f => apply(pure(f));
const asyncSum = lift(sum);
const result = asyncSum(async1, async2, async3, async4);

请注意，reduce 等效于 arrFold 。因此，lift等同于varLiftM .