通过map`的'行为对使用`new`创建的阵列困惑 [英] Confused by behavior of `map` on arrays created using `new`

问题描述

我对地图平的结果与上创建一个数组混淆新：

I am confused by the results of mapping over an array created with new:

function returnsFourteen() {
    return 14;
}

var a = new Array(4);
> [undefined x 4] in Chrome, [, , , ,] in Firefox
a.map(returnsFourteen);
> [undefined x 4] in Chrome, [, , , ,] in Firefox

var b = [undefined, undefined, undefined, undefined];
> [undefined, undefined, undefined, undefined]
b.map(returnsFourteen);
> [14, 14, 14, 14]

我的预期 a.map（returnsFourteen）返回 [14，14，14，14] （中同 b.map（returnsFourteen），因为根据的对数组MDN页面：

I expected a.map(returnsFourteen) to return [14, 14, 14, 14] (the same as b.map(returnsFourteen), because according to the MDN page on arrays:

如果传递给Array构造函数的唯一参数是一个整数
  2 ** 32-1（含）之间的0，并创建一个新的JavaScript数组
  与一些元素。

If the only argument passed to the Array constructor is an integer between 0 and 2**32-1 (inclusive), a new JavaScript array is created with that number of elements.

我间preT这意味着 A 应该有4个元素。

I interpret that to mean that a should have 4 elements.

我是什么在这里失踪？

推荐答案

在创建这样一个数组：

var arr1 = new Array( 4 );

你有 4 的长度的数组，但拥有的没有的元素。这就是为什么地图不转换阵列 - 数组有没有被转化元素

you get an array that has a length of 4, but that has no elements. That's why map doesn't tranform the array - the array has no elements to be transformed.

在另一方面，如果你这样做：

On the other hand, if you do:

var arr2 = [ undefined, undefined, undefined, undefined ];

您获取和数组，也有 4 的长度，但确实的有4个元素。

you get and array that also has a length of 4, but that does have 4 elements.

注意，无元素，具有元件，其值未定义之间的差异。不幸的是，属性访问前pression将评估在两种情况下，未定义价值，所以：

Notice the difference between having no elements, and having elements which values are undefined. Unfortunately, the property accessor expression will evaluate to the undefined value in both cases, so:

arr1[0] // undefined
arr2[0] // undefined

然而，区分这两个阵列的一种方式：

However, there is a way to differentiate these two arrays:

'0' in arr1 // false
'0' in arr2 // true

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