被`new`创建的数组上的`map`的行为所困惑 [英] Confused by behavior of `map` on arrays created using `new`

问题描述

我对映射的结果感到困惑，使用新建 c $ c创建的数组：

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]

我预计 a.map returnFourteen）返回 [14，14，14，14] （与相同b.map（returnsFourteen），因为根据阵列上的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
，介于0和2之间** 32-1（含），一个新的JavaScript数组被创建
与这个元素数量。

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.

我解释说，这意味着 a 应该有4个元素。

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

我在这里缺少什么？

推荐答案

当你创建一个这样的数组时：

When you create an array like so:

var arr1 = new Array( 4 );

你得到一个长度为 4 ，但是那个没有元素，这就是为什么 map 不会转换数组 - 数组没有要转换的元素。

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.

另一方面，如果你这样做：

On the other hand, if you do:

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

你得到的数组也有一个长度 4 ，但是 具有4个元素。

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

注意没有元素之间的区别，并且元素的值为未定义。不幸的是，在这两种情况下，属性访问器表达式将评估为 undefined 值，因此：

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

但是，有一种区分这两个数组的方法：

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

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

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