第 166 题：闭包 JS 基础 编程题（字节）

Question

var foo = function(...args) { // 要求实现函数体}
var f1 = foo(1,2,3); f1.getValue(); // 6 输出是参数的和
var f2 = foo(1)(2,3); f2.getValue(); // 6
var f3 = foo(1)(2)(3)(4); f3.getValue(); // 10

Answer 1

Tip:

1. 返回值是一个函数
2. 动态参数，拼接参数，getValue 统一计算

function foo(...args) {
  let addArgs = [...args];
  function fn(...innerArgs) {
    addArgs = [...addArgs, ...innerArgs];
    return fn;
  }
  fn.getValue = function () {
    return addArgs.reduce((pre, cur) => {
      return pre + cur;
    }, 0);
  };
  return fn;
}

const sum = foo(1, 2, 3)(2)(3).getValue();
// const sum = foo(1, 2, 3).getValue();
console.log("sum", sum);

更多解析

Answer 2

const foo = (...args) => {
  const _add = (...args1) => {
    return foo(...args, ...args1);
  }

  _add.getValue = () => args.reduce((acc, cur) => acc + cur);

  return _add;
}

Answer 3

function foo(...args) {
  const sum = args.reduce((acc, v) => acc + v, 0);

  const hof = function (...args) {
    return foo(sum, ...args);
  };
  hof.getValue = () => sum;

  return hof;
}

Answer 4

const foo = function() {
    let args = Array.prototype.slice.call(arguments);
    const target = function() {
        args = args.concat(Array.prototype.slice.call(arguments));
        return foo(...args);
    };
    target.getValue = () => args.reduce((x, y) => x + y, 0);
    return target;
};

Answer 5

@Ishmael-Yoko 不错，大概跟面试官想要的一致。

Answer 6

function foo(...args) {
  const target = (...arg1s) => foo(...[...args, ...arg1s])
  target.getValue = () => args.reduce((p, n) => p+ n, 0)
  return target
}

Answer 7

function foo(...args) {
  foo.value = args.reduce((total, cur) => total + cur, (foo.value || 0))
  return foo
}

foo.getValue = function () {
  var value = this.value;
  this.value = 0;
  return value
}

Answer 8

Function.prototype.getValue = function () {
    let num = this.arr.reduce((x, y) => (x + y))
    this.arr = null
    return num
  }
  var foo = function (...args) {
    if (!Array.isArray(foo.arr)) {
      foo.arr = []
    }
    foo.arr.push(...args)
    return foo;
  }