如何为应用计算实现协程？

Question

这是一个避免嵌套模式的协程，例如 (chain(m) (chain(...)) 用于单子计算：

const some = x => none => some => some(x);
const none = none => some => none;

const option = none => some => tx => tx(none) (some);
const id = x => x;

const of = some;
const chain = fm => m => none => some => m(none) (x => fm(x) (none) (some));

const doM = (chain, of) => gf => {
  const it = gf();

  const loop = ({done, value}) =>
    done
      ? of(value)
      : chain(x => loop(it.next(x))) (value);

  return loop(it.next());
};

const tx = some(4),
  ty = some(5),
  tz = none;

const data = doM(chain, of) (function*() {
  const x = yield tx,
    y = yield ty,
    z = yield tz;

  return x + y + z;
});

console.log(
  option(0) (id) (data)); // 0

但我无法为应用计算实现等效的协程：

const some = x => none => some => some(x);
const none = none => some => none;

const option = none => some => tx => tx(none) (some);
const id = x => x;

const of = some;
const map = f => t => none => some => t(none) (x => some(f(x)));
const ap = tf => t => none => some => tf(none) (f => t(none) (x => some(f(x))));

const doA = (ap, of) => gf => {
  const it = gf();

  const loop = ({done, value}, initial) =>
    done
      ? value
      : ap(of(x => loop(it.next(x)))) (value);

  return loop(it.next());
};

const tx = some(4),
  ty = some(5),
  tz = none;

const data = doA(ap, of) (function*() {
  const x = yield tx,
    y = yield ty,
    z = yield tz;

  return x + y + z;
});

console.log(
  option(0) (id) (data)); // none => some => ...

这应该有效，但它没有。额外的功能包装从何而来？我想我在这里的递归有点迷失了。

顺便说一句，我知道这仅适用于确定性仿函数/单子。

Answer 1

我无法为应用计算实现等效的协程

是的，因为生成器函数是一元的，而不仅仅是应用程序。 yield 表达式的操作数可以取决于前一个 yield 表达式的结果 - 这是 monad 的特征。

额外的功能包装从何而来？我想我在这里有点迷路了。

你正在做ap(of(…))(…) - 根据Applicative laws，这相当于map(…)(…)。与第一个 sn-p 中的 chain 调用相比，这不会对结果进行任何解包，因此您会得到一个嵌套的 maybe 类型（在您的实现中，它被编码为一个函数）。