Extending String native type in Typescript

Question

I would like to extend the String type in Typescript. Here's my code :

export class Str extends String{
    constructor(str: string) {
        super(str);
    }
}

The problem is, after the call of the String constructor using super(str), the value of my new type stills empty.

I have the same problem with another type extending the native Array type.

MartyIX · Accepted Answer · 2016-01-28 16:39:35Z

3

Your code is transpiled to:

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 Str = (function (_super) {
    __extends(Str, _super);
    function Str(str) {
        _super.call(this, str);
    }
    return Str;
})(String);

And the problem is that you no longer deal with the original native type that has special properties (like + operator on two strings). So your extended type - even if you would have succeded - would be severely limited and confusing for others.

Details and gotchas can be found Axel Rauschmayer's article: http://www.2ality.com/2013/03/subclassing-builtins-es6.html

answered Jan 28, 2016 at 16:39

MartyIX

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3 Comments

Martin Lezer Over a year ago

Hi Martin. Thanks for your answer. Does that mean that Typescript is not able to extend native type? If so the only way is to write Javascript code?

MartyIX Over a year ago

Well, it's not built TypeScript to handle these special cases, so extends keywords does not work out of box. The problem is that I don't think you can fully create your own string with properties of the built-in type. So you are left with the option to modify built-in type and that is a terrible idea.

Martin Lezer Over a year ago

OK thanks. Actually I already have a workaround. I was just trying to refactor my code to make it cleaner. That's too bad I was not able to do this with Typescript.

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