javascript - 反转 JavaScript 中默认参数的顺序

Question

我有以下递归compose函数:

const compose = (f, n = 1) => n > 1 ?
    compose(compose(f), n - 1) :
    g => x => f(g(x));

const length = a => a.length;

const filter = p => a => a.filter(p);

const countWhere = compose(length, 2)(filter);

const odd = n => n % 2 === 1;

console.log(countWhere(odd)([1,2,3,4,5,6,7,8,9])); // 5

现在，我想做的是翻转 compose 的参数，以便默认参数位于第一个:

const compose = (n = 1, f) => n > 1 ? // wishful thinking
    compose(n - 1, compose(f)) : // compose(f) is the same as compose(1, f)
    g => x => f(g(x));

const length = a => a.length;

const filter = p => a => a.filter(p);

const countWhere = compose(2, length)(filter); // I want to call it like this

const odd = n => n % 2 === 1;

console.log(countWhere(odd)([1,2,3,4,5,6,7,8,9])); // 5

编写默认参数优先的函数的最优雅的方法是什么？

<小时/>

编辑:我实际上想创建 map和 ap各种函数的方法，这样我就可以写:

const length = a => a.length;

const filter = p => a => a.filter(p);

const countWhere = length.map(2, filter); // length <$> filter

const pair = x => y => [x, y];

const add = x => y => x + y;

const mul = x => y => x * y;

const addAndMul = pair.map(2, add).ap(2, mul); // (,) <$> (+) <*> (*)

因此，我不想像 Bergi 在他的回答中建议的那样柯里化(Currying)方法。

有关更多信息，请阅读:Is implicit wrapping and unwrapping of newtypes in Haskell a sound idea?

Answer 1

我建议不要重载您的函数或使用默认参数:

const compose = n => f => n > 1
  ? compose(n - 1)(composeOne(f))
  : g => x => f(g(x));
const composeOne = compose(1);

在这种情况下，您也可以将其内联，因为看起来 composeOne 不会在其他任何地方调用:

const compose = n => f => n > 1
  ? compose(n - 1)(compose(1)(f))
  : g => x => f(g(x));

或者甚至根本不进行递归调用，但始终创建​​ g => x => ... lambda 并有条件地对其进行转换:

const compose = n => f => {
  const l = g => x => f(g(x));
  return n > 1 ? compose(n - 1)(l) : l;
};
// same without temporary variables:
const id = x => x;
const compose = n => f => (n > 1 ? compose(n-1) : id)(g => x => f(g(x)))