--- id: 9d7123c8c441eeafaeb5bdef challengeType: 1 forumTopicId: 301236 title: 使用 slice 而不是 splice 从数组中移除元素 --- ## Description
使用数组时经常遇到要删除一些元素并保留数组剩余部分的情况。为此,JavaScript 提供了splice方法,它接收两个参数:从哪里开始删除项目的索引,和要删除的项目数。如果没有提供第二个参数,默认情况下是移除到结尾的元素。但splice方法会改变调用它的原始数组。举个例子: ```js var cities = ["Chicago", "Delhi", "Islamabad", "London", "Berlin"]; cities.splice(3, 1); // Returns "London" and deletes it from the cities array // cities is now ["Chicago", "Delhi", "Islamabad", "Berlin"] ``` 正如我们在上一次挑战中看到的那样,slice方法不会改变原始数组,而是返回一个可以保存到变量中的新数组。回想一下,slice方法接收两个参数,从开始索引开始选取到结束(不包括该元素),并在新数组中返回这些元素。使用slice方法替代splice有助于避免数组变化产生的副作用。
## Instructions
slice代替splice重写nonMutatingSplice函数。将cities数组长度限制为3,并返回一个仅包含前 3 项的新数组。 不要改变提供给函数的原始数组。
## Tests
```yml tests: - text: 应该使用slice方法。 testString: assert(code.match(/\.slice/g)); - text: 不能使用splice方法。 testString: assert(!code.match(/\.splice/g)); - text: 不能改变inputCities数组。 testString: assert(JSON.stringify(inputCities) === JSON.stringify(["Chicago", "Delhi", "Islamabad", "London", "Berlin"])); - text: "nonMutatingSplice(['Chicago', 'Delhi', 'Islamabad', 'London', 'Berlin'])应返回['Chicago', 'Delhi', 'Islamabad']。" testString: assert(JSON.stringify(nonMutatingSplice(["Chicago", "Delhi", "Islamabad", "London", "Berlin"])) === JSON.stringify(["Chicago", "Delhi", "Islamabad"])); ```
## Challenge Seed
```js function nonMutatingSplice(cities) { // Add your code below this line return cities.splice(3); // Add your code above this line } var inputCities = ["Chicago", "Delhi", "Islamabad", "London", "Berlin"]; nonMutatingSplice(inputCities); ```
## Solution
```js function nonMutatingSplice(cities) { // Add your code below this line return cities.slice(0,3); // Add your code above this line } var inputCities = ["Chicago", "Delhi", "Islamabad", "London", "Berlin"]; nonMutatingSplice(inputCities); ```