javascript - 堆栈安全的相互递归，不会在调用端泄露实现细节

Question

我概括了clojure的loop/recur蹦床，以便它与间接递归一起工作：

const trampoline = f => (...args) => {
  let acc = f(...args);

  while (acc && acc.type === recur) {
    let [f, ...args_] = acc.args;
    acc = f(...args_);
  }

  return acc;
};

const recur = (...args) =>
  ({type: recur, args});

const even = n =>
  n === 0
    ? true
    : recur(odd, n - 1);

const odd = n =>
  n === 0
    ? false
    : recur(even, n - 1);


console.log(
  trampoline(even) (1e5 + 1)); // false

但是，我必须在调用方显式调用蹦床。有没有办法让它再次隐含，就像loop/一样recur？

顺便说一句，这里是loop/ recur：

const loop = f => {
  let acc = f();

  while (acc && acc.type === recur)
    acc = f(...acc.args);

  return acc;
};


const recur = (...args) =>
  ({type: recur, args});

Answer 1

显然，由于您希望调用蹦床，因此不能完全跳过它。最简单的事情就是将这些蹦床调用包装在您想要的 API 中，可能是这样的：

// Utility code
const trampoline = f => (...args) => {
  let acc = f(...args);

  while (acc && acc.type === recur) {
    let [f, ...args_] = acc.args;
    acc = f(...args_);
  }

  return acc;
};

const recur = (...args) =>
  ({type: recur, args});

// Private implementation
const _even = n =>
  n === 0
    ? true
    : recur(_odd, n - 1);

const _odd = n =>
  n === 0
    ? false
    : recur(_even, n - 1);

// Public API
const even = trampoline(_even);

const odd = trampoline(_odd);

// Demo code
console.log(
  even (1e5 + 1)); // false

console.log(
  odd (1e5 + 1)); // true