双语面试：实现一个柯里化函数

面试题 Interview Question

实现一个函数 curry，可以把任意多参数函数转换成支持多次调用的柯里化函数。
Implement a function curry that transforms any multi-argument function into a curried version that supports multiple chained calls.

要求 Requirements

1. curry(fn) 接收一个多参数函数 fn，返回一个可以逐个接收参数的函数
   curry(fn) takes a function fn with multiple parameters and returns a new function that can take arguments one by one
2. 支持如下调用方式：
   The resulting function should support the following usage:

const add = (a, b, c) => a + b + c;
const curriedAdd = curry(add);

curriedAdd(1)(2)(3); // 输出 6 → Outputs: 6
curriedAdd(1, 2)(3); // 输出 6 → Outputs: 6
curriedAdd(1)(2, 3); // 输出 6 → Outputs: 6

3. 不使用全局变量，不使用 this，保持纯函数风格
   Do not use global variables or this, keep the implementation in a pure functional style

参考答案 Reference Solution

function curry(fn) {
  return function curried(...args) {
    if (args.length >= fn.length) {
      return fn(...args);
    } else {
      return (...nextArgs) => curried(...args, ...nextArgs);
    }
  };
}

示例 Example

function multiply(a, b, c) {
  return a * b * c;
}

const curriedMultiply = curry(multiply);

console.log(curriedMultiply(2)(3)(4));    // 输出 Output: 24
console.log(curriedMultiply(2, 3)(4));    // 输出 Output: 24
console.log(curriedMultiply(2)(3, 4));    // 输出 Output: 24

面试考察点 Interview Focus

• 对函数柯里化与偏函数的理解
  Understanding of currying vs. partial application

• 函数参数处理技巧（rest/spread 运算符）
  Handling function arguments with rest/spread syntax

• 闭包和函数式编程思想掌握程度
  Mastery of closures and functional programming concepts