数组上的 JavaScript 头和尾没有变化
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JavaScript head and tail on array without mutation
提问by Knows Not Much
I found about JavaScript array operations Unshift, shift, push, pop
我发现了 JavaScript 数组操作 Unshift、shift、push、pop
However all these operations mutate the array.
然而，所有这些操作都会改变数组。
Is there a way I could use these functions without causing mutation on the original data?
有没有一种方法可以使用这些函数而不会导致原始数据发生突变？
Somehow I feel that reading the data should not cause mutation.
不知怎的，我觉得读取数据不应该引起突变。
回答by Davin Tryon
You can use:
您可以使用：
var head = arr[0];
var tail = arr.slice(1);
Or in ES6:
或者在 ES6 中：
const [head, ...tail] = arr;
回答by Suhail Mumtaz Awan
I would do using ES6 as below
我会使用 ES6 如下
const head = ([h]) => h;
const tail = ([, ...t]) => t;
const arr = [1,2,3,4,5];
alert(head(arr));
回答by Bamieh
The accepted answer is good, but for a more functional approach:
接受的答案很好，但对于更实用的方法：
Heads
头
I'd use find, which returns the first element that returns a true-thy value in its predicate.
我会使用 find，它返回第一个在其谓词中返回真值的元素。
If you are sure your values are true-thy, you can write it as follows:
如果你确定你的价值观是真实的，你可以这样写：
arr.find(Boolean)
If you want the first value, regardless of its value, you can write it as follows:
如果你想要第一个值，不管它的值是多少，你可以这样写：
arr.find(_ => true)
Tails
尾巴
Just as the others state, use slice
正如其他人所说，使用 slice
arr.slice(1);
