3

我有一个可变参数提升函数，它允许扁平的单子链没有深度嵌套的函数组合：

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，但我想通过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类型无关，

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

因为行中的代数A会为每次迭代返回 a Task，而不是部分应用的函数。

如何用折叠替换非尾递归？

这是当前递归实现的一个工作示例：

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] 根据评论中的要求，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;
};
4

2 回答 2

1

那个of呼唤arrFold似乎有点不合时宜。

我不确定您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)))

变成

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

用左折叠，你可以写

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

但是您需要注意，这会构建与正确折叠不同的调用树：

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（已经应用了延续）

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)))))

根据单子定律，它们应该评估相同，但在实践中，一个可能比另一个更有效。

于 2019-06-02T19:26:54.940 回答
1

对于这个特定的用例，您不需要 monad 的全部功能。您只需要应用函子：

// 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...");

如果你愿意，你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 curry 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这相当于arrFold。因此，lift等价于varLiftM。

于 2019-06-03T11:49:07.623 回答