2.9 KiB
2.9 KiB
id, challengeType, forumTopicId, title
id | challengeType | forumTopicId | title |
---|---|---|---|
9d7123c8c441eeafaeb5bdef | 1 | 301236 | 使用 slice 而不是 splice 从数组中移除元素 |
Description
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:
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
tests:
- text: 应该使用<code>slice</code>方法。
testString: assert(code.match(/\.slice/g));
- text: 不能使用<code>splice</code>方法。
testString: assert(!code.match(/\.splice/g));
- text: 不能改变<code>inputCities</code>数组。
testString: assert(JSON.stringify(inputCities) === JSON.stringify(["Chicago", "Delhi", "Islamabad", "London", "Berlin"]));
- text: "<code>nonMutatingSplice(['Chicago', 'Delhi', 'Islamabad', 'London', 'Berlin'])</code>应返回<code>['Chicago', 'Delhi', 'Islamabad']</code>。"
testString: assert(JSON.stringify(nonMutatingSplice(["Chicago", "Delhi", "Islamabad", "London", "Berlin"])) === JSON.stringify(["Chicago", "Delhi", "Islamabad"]));
Challenge Seed
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
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);