子类化JavaScript数组 - 不传递给数组的构造函数参数 [英] Subclassing JavaScript arrays - constructor arguments not passing to array

问题描述

我想继承javascript数组并将构造函数参数传递给数组。我拥有的是：

I want to subclass a javascript Array and pass constructor arguments to the array. What I have is:

function SubArray(arguments) {
    Array.apply(this, arguments);
}

SubArray.prototype = Array.prototype;

测试并不表示参数传递给数组

Test does not indicate that arguments are being passed to the Array

var x = new SubArray("One", "Two", "Three");
// Object[]
// x.length = 0

当我用数组做这个时我得到这个

Whereas when I do this with an Array I get this

var x = new Array("One", "Two", "Three");
// Object["One", "Two", "Three"]
// x.length = 3

我做错了什么？

推荐答案

ES6

在ES6中，您可以使用类语法扩展 Array

class SubArray extends Array {
    constructor() {
        super(...arguments);
    }
    first() {
        return this[0];
    }
}

length 将作为预期

var sa = new SubArray('foo', 'bar');
sa[2] = 'baz';
sa.length; // 3
sa.first(); // "foo"
sa instanceof SubArray; // true

---

Pre-ES5

直到并包括ES5，没有办法以通常的构造函数原型方式干净地扩展 Array 而不会丢失功能，而是你必须向Array实例添加属性

---

Pre-ES5

Up to and including ES5 there is no way to cleanly extend Array in the usual constructor-prototype way without losing functionality, instead you have to add properties to an Array instance

function SubArray() {
    var arr = Array.apply(null, arguments);
    arr.first = function () {return this[0];};
    return arr;
}

使用它不再需要 new 从函数返回 Object ， instanceof 将无法确定 SubArray 。

Using this no longer requires new as an Object is returned from the function, and instanceof will not be able to determine SubArray.

var sa = SubArray('foo', 'bar');
sa[2] = 'baz';
sa.length; // 3
sa.first(); // "foo"
sa instanceof SubArray; // false -- this gets broken

---

ES5（慢速）

在行为比速度更重要的ES5环境中，您可以使用 Object.setPrototypeOf （请参阅MDN警告） ，这是有点黑客，看起来像

function SubArray() {
    var arr = Array.apply(null, arguments);
    Object.setPrototypeOf(arr, SubArray.prototype);
    return arr;
}
SubArray.prototype = Object.create(Array.prototype);
SubArray.prototype.first = function () {return this[0];};

再次，不再需要 new 但这次实例的行为与正常情况下的预期完全相同。

Again, new is no-longer required but this time the behaviour of instances is exactly as would be expected if it was extended normally.

var sa = SubArray('foo', 'bar');
sa[2] = 'baz';
sa.length; // 3
sa.first(); // "foo"
sa instanceof SubArray; // true

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