Question

我正在尝试功能List类型和结构共享。由于Javascript没有Tail Recursive Modulo Cons优化，我们不能像这样编写List个组合器，因为它们不是堆栈安全的：

＆＃13;

const list =
  [1, [2, [3, [4, [5, []]]]]];


const take = n => ([head, tail]) =>
  n === 0 ? []
    : head === undefined ? []
    : [head, take(n - 1) (tail)];


console.log(
  take(3) (list) // [1, [2, [3, []]]]
);

＆＃13;

现在我尝试递归地实现take尾部，这样我就可以依赖TCO（在Ecmascript中仍然是一个未解决的Promise）或使用蹦床（在示例中省略以保持简单）：

＆＃13;

const list =
  [1, [2, [3, [4, [5, []]]]]];


const safeTake = n => list => {
  const aux = (n, acc, [head, tail]) => n === 0 ? acc
    : head === undefined ? acc
    : aux(n - 1, [head, acc], tail);

  return aux(n, [], list);
};


console.log(
  safeTake(3) (list) // [3, [2, [1, []]]]
);

＆＃13;

这有效但新创建的列表的顺序相反。如何以纯粹的功能性方式解决这个问题？

Answer 1

防止列表反转的一种方法是使用延续传递样式。现在就把它放在你选择的蹦床上......

＆＃13;

const None =
  Symbol ()

const identity = x =>
  x

const safeTake = (n, [ head = None, tail ], cont = identity) =>
  head === None || n === 0
    ? cont ([])
    : safeTake (n - 1, tail, answer => cont ([ head, answer ]))

const list =
  [ 1, [ 2, [ 3, [ 4, [ 5, [] ] ] ] ] ]

console.log (safeTake (3, list))
// [ 1, [ 2, [ 3, [] ] ] ]

＆＃13;

这是在蹦床上

const None =
  Symbol ()

const identity = x =>
  x

const call = (f, ...values) =>
  ({ tag: call, f, values })

const trampoline = acc =>
{
  while (acc && acc.tag === call)
    acc = acc.f (...acc.values)
  return acc
}

const safeTake = (n = 0, xs = []) =>
{
  const aux = (n, [ head = None, tail ], cont) =>
    head === None || n === 0
      ? call (cont, [])
      : call (aux, n - 1, tail, answer =>
          call (cont, [ head, answer ]))
  return trampoline (aux (n, xs, identity))
}

const list =
  [ 1, [ 2, [ 3, [ 4, [ 5, [] ] ] ] ] ]

console.log (safeTake (3, list))
// [ 1, [ 2, [ 3, [] ] ] ]

Answer 2

Laziness免费为您提供尾递归模数。因此，显而易见的解决方案是使用thunk。但是，我们不只是想要任何一种thunk。我们想要weak head normal form中的表达式。在JavaScript中，我们可以使用lazy getters实现此目的，如下所示：

const cons = (head, tail) => ({ head, tail });

const list = cons(1, cons(2, cons(3, cons(4, cons(5, null)))));

const take = n => n === 0 ? xs => null : xs => xs && {
    head: xs.head,
    get tail() {
        delete this.tail;
        return this.tail = take(n - 1)(xs.tail);
    }
};

console.log(take(3)(list));

使用懒惰的getter有很多好处：

正常属性和延迟属性以相同的方式使用。
您可以使用它来创建无限的数据结构。
你不必担心炸毁堆栈。

希望有所帮助。

如何防止尾递归函数反转List的顺序？

2 个答案: