使用n个参数在javascript中进行函数的currying

Question

如果f ::(a,b) - > c,我们可以定义curry(f)如下:

咖喱(f)::((a,b) - > c) - > a - > b - > c

const curry = f => a => b => f(a, b);
const sum = curry((num1, num2) => num1 + num2);
console.log(sum(2)(3)); //5

我们如何实现带有n个参数的函数的通用咖喱函数？

Answer 1

如果我理解正确,我认为这是使用ES6的方法:

const curry = f => {
  const nargs = f.length;
  const vargs = [];
  const curried = (...args) => vargs.push(...args) >= nargs
    ? f(...vargs.slice(0, nargs))
    : curried;

  return curried;
};

const fn2 = curry((a, b) => a + b);
const fn3 = curry((a, b, c) => a * (b + c));
const fn4 = curry((a, b, c, d) => Math.pow(a, b * (c + d)));

console.log(fn2(1)(2)); // 1 + 2
console.log(fn3(2)(3)(4)); // 2 * (3 + 4)
console.log(fn4(2)(1, 3)(4)); // 2 ^ (1 * (3 + 4))

如果你想在ES5中这样做,这里有一个稍微冗长的方法:

function curry (f) {
  var nargs = f.length;
  var vargs = [];

  return function curried () {
    return vargs.push.apply(vargs, arguments) >= nargs
      ? f.apply(undefined, vargs.slice(0, nargs))
      : curried;
  };
}

var fn2 = curry(function (a, b) {
  return a + b;
});
var fn3 = curry(function (a, b, c) {
  return a * (b + c);
});
var fn4 = curry(function (a, b, c, d) {
  return Math.pow(a, b * (c + d));
});

console.log(fn2(1)(2)); // 1 + 2
console.log(fn3(2)(3)(4)); // 2 * (3 + 4)
console.log(fn4(2)(1, 3)(4)); // 2 ^ (1 * (3 + 4))