Class extended from built-in Array in Typescript 1.6.2 does not update length while using [] operator

Question

it should be possible to extend build-in types in ts 1.6 as I read here:

TypeScript 1.6 adds support for classes extending arbitrary expression that comp

Answer 1

Your code translates to this JavaScript code:

var __extends = (this && this.__extends) || function (d, b) {
    for (var p in b) if (b.hasOwnProperty(p)) d[p] = b[p];
    function __() { this.constructor = d; }
    d.prototype = b === null ? Object.create(b) : (__.prototype = b.prototype, new __());
};
var ExtendedArray = (function (_super) {
    __extends(ExtendedArray, _super);
    function ExtendedArray() {
        _super.apply(this, arguments);
    }
    return ExtendedArray;
})(Array);
var buildinArray = new Array();
var extendedArray = new ExtendedArray();
buildinArray.push("A");
console.log(buildinArray.length); // 1 - OK
buildinArray[2] = "B";
console.log(buildinArray.length); // 3 - OK
extendedArray.push("A");
console.log(extendedArray.length); // 1 - OK
extendedArray[2] = "B";
console.log(extendedArray.length); // 1 - FAIL
console.dir(extendedArray); // both values, but wrong length

[Playground]

The problem is that the bracket notation is not copied as part of copying of Array prototype:

In JavaScript, we can sub-class native data types by extending the native prototypes. This works perfectly with the native String object; but, when it comes to native Arrays, things don't work quite so nicely. If we extend the Array prototype, we inherit the native array functions; but, we no longer have the ability to use bracket notation to set and get indexed values within the given array. Sure, we can use push() and pop() to overcome this limitation; but, if we want to keep the bracket notation feature functional, we have to build on top of an existing array instance rather than truly sub-classing the Array object.

[Source]

There is an in-depth discussion and solutions here by Dr. Axel Rauschmayer.