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freeCodeCamp/curriculum/challenges/chinese/08-coding-interview-prep/rosetta-code/accumulator-factory.chinese.md
Kristofer Koishigawa b3213fc892 fix(i18n): chinese test suite (#38220)
* fix: Chinese test suite

Add localeTiltes, descriptions, and adjust test text and testStrings to get the automated test suite working.

* fix: ran script, updated testStrings and solutions
2020-03-03 18:49:47 +05:30

1.4 KiB
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title, id, challengeType, videoUrl, localeTitle
title id challengeType videoUrl localeTitle
Accumulator factory 594810f028c0303b75339ace 5 蓄能器工厂

Description

创建一个带有单个(数字)参数的函数,并返回另一个作为累加器的函数。返回的累加器函数又接受一个数字参数,并返回到目前为止传递给该累加器的所有数值的总和(包括创建累加器时传递的初始值)。

规则:

不要使用全局变量。

暗示:

闭包可以保存外部状态。

Instructions

Tests

tests:
  - text: <code>accumulator</code>是一个功能。
    testString: assert(typeof accumulator === 'function');
  - text: <code>accumulator(0)</code>应该返回一个函数。
    testString: assert(typeof accumulator(0) === 'function');
  - text: <code>accumulator(0)(2)</code>应该返回一个数字。
    testString: assert(typeof accumulator(0)(2) === 'number');
  - text: '传递值3-4,1.5和5应返回5.5。'
    testString: assert(testFn(5) === 5.5);

Challenge Seed

function accumulator (sum) {
  // Good luck!
}

After Test

console.info('after the test');

Solution

// solution required