实现一个 sum() 使得如下判断成立。

示例

const sum1 = sum(1)
const _sum = sum()

sum1 == 1 // true
_sum == 0 // true

sum1(2) == 3 // true
sum1(3) == 4 // true

sum(1)(2)(3) == 6 // true
sum(5)(-1)(2) == 6 // true
sum(1)(2)() == 3 // true
sum(1)()(2) == 3 // true

• 注意是 == 而不是 ===
• 可以连续调用
• 如果不传入参数，默认为 0

题目模版

sum.js

/**
 * @param {number} v 传入的值
 */
export default function sum(v = 0) {

}

sum.ts

/**
 * @param v 传入的值
 */
function sum(v = 0) {}

export default sum

types.ts

// 求和函数的返回类型，既是函数又可以转换为数字
export interface SumWrapper {
  (newArg?: number): SumWrapper
  [Symbol.toPrimitive]: () => number
}

// 主求和函数接口
export interface SumFunction {
  (v?: number): SumWrapper
}

测试代码

sum.spec.js

import { describe, expect, it } from 'vitest'
import sum from './sum'

describe('04.21--default.实现一个sum()方法', () => {
  it('应在传入一个参数的情况下正常运行', () => {
    const sum1 = sum(1)

    expect(sum1 == 1).toBeTruthy()
  })

  it('应在不传入参数的情况下正常运行', () => {
    const sum1 = sum()

    expect(sum1 == 0).toBeTruthy()
  })

  it('应在连续调用多次的情况下正常运行', () => {
    const sum1 = sum(1)(2)(3)

    expect(sum1 == 6).toBeTruthy()
  })

  it('应不影响其他函数调用', () => {
    const sum1 = sum(1)

    expect(sum1(2) == 3).toBeTruthy()
    expect(sum1(3) == 4).toBeTruthy()
  })

  it('应在多种情况下正常运行', () => {
    const sum1 = sum(1)

    expect(sum1(2)()(3) == 6).toBeTruthy()
  })

  it('应在传入负数的情况下正常运行', () => {
    const sum1 = sum(-1)(-2)(-3)

    expect(sum1 == -6).toBeTruthy()
  })

  it('应在传入小数的情况下正常运行', () => {
    const sum1 = sum(1.5)(2.5)(3.5)

    expect(sum1 == 7.5).toBeTruthy()
  })

  it('应在混合正数和负数的情况下正常运行', () => {
    const sum1 = sum(5)(-3)(2)

    expect(sum1 == 4).toBeTruthy()
  })

  it('应在连续调用后返回正确的结果', () => {
    const sum1 = sum(10)
    const sum2 = sum1(20)
    const sum3 = sum2(30)

    expect(sum3 == 60).toBeTruthy()
  })

  it('应在多次调用后保持独立的状态', () => {
    const sum1 = sum(1)
    const sum2 = sum1(2)
    const sum3 = sum(3)

    expect(sum2(3) == 6).toBeTruthy()
    expect(sum3(4) == 7).toBeTruthy()
  })
})

sum.spec.ts

/* eslint-disable ts/ban-ts-comment */
// @ts-nocheck
import { describe, expect, it } from 'vitest'
import sum from './sum'

describe('04.21--default.实现一个sum()方法', () => {
  it('应在传入一个参数的情况下正常运行', () => {
    const sum1 = sum(1)

    expect(+sum1 == 1).toBeTruthy()
  })

  it('应在不传入参数的情况下正常运行', () => {
    const sum1 = sum()

    expect(+sum1 == 0).toBeTruthy()
  })

  it('应在连续调用多次的情况下正常运行', () => {
    const sum1 = sum(1)(2)(3)

    expect(+sum1 == 6).toBeTruthy()
  })

  it('应不影响其他函数调用', () => {
    const sum1 = sum(1)

    expect(+sum1(2) == 3).toBeTruthy()
    expect(+sum1(3) == 4).toBeTruthy()
  })

  it('应在多种情况下正常运行', () => {
    const sum1 = sum(1)

    expect(+sum1(2)()(3) == 6).toBeTruthy()
  })

  it('应在传入负数的情况下正常运行', () => {
    const sum1 = sum(-1)(-2)(-3)

    expect(+sum1 == -6).toBeTruthy()
  })

  it('应在传入小数的情况下正常运行', () => {
    const sum1 = sum(1.5)(2.5)(3.5)

    expect(+sum1 == 7.5).toBeTruthy()
  })

  it('应在混合正数和负数的情况下正常运行', () => {
    const sum1 = sum(5)(-3)(2)

    expect(+sum1 == 4).toBeTruthy()
  })

  it('应在连续调用后返回正确的结果', () => {
    const sum1 = sum(10)
    const sum2 = sum1(20)
    const sum3 = sum2(30)

    expect(+sum3 == 60).toBeTruthy()
  })

  it('应在多次调用后保持独立的状态', () => {
    const sum1 = sum(1)
    const sum2 = sum1(2)
    const sum3 = sum(3)

    expect(+sum2(3) == 6).toBeTruthy()
    expect(+sum3(4) == 7).toBeTruthy()
  })
})

答案

类型	路径
JS 版本	problems/Day 04/answer.js
TS 版本	problems/Day 04/ts/answer.ts
Review	04.md