Unexpected behaviour

[SimonCsoma]

TypeScript Version: 1.8.0
Code

class Animal {
    constructor() { 
        this.move()
    }
    move(distanceInMeters: number = 0) {
        console.log(distanceInMeters);
    }
}

class Snake extends Animal {
    public variable = 1;
    constructor() { 
        super();

    }
    move(distanceInMeters = 5) {
        console.log(this.variable);
        super.move(distanceInMeters);
    }
}

Expected behavior:
Class variables should alway be accessible and set and should be compiled like this:

    function Snake() {
        this.variable = 1;
        _super.call(this);
    }

Entire compilation:

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 Animal = (function () {
    function Animal() {
        this.move();
    }
    Animal.prototype.move = function (distanceInMeters) {
        if (distanceInMeters === void 0) { distanceInMeters = 0; }
        console.log(distanceInMeters);
    };
    return Animal;
}());
var Snake = (function (_super) {
    __extends(Snake, _super);
    function Snake() {
        this.variable = 1;
        _super.call(this);
    }
    Snake.prototype.move = function (distanceInMeters) {
        if (distanceInMeters === void 0) { distanceInMeters = 5; }
        console.log(this.variable);
        _super.prototype.move.call(this, distanceInMeters);
    };
    return Snake;
}(Animal));

Actual behavior:
This causes this.variable in the move function of the snake class to always be undefined when constructing a new class.

    function Snake() {
        _super.call(this);
        this.variable = 1;
    }

Entire compilation:

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 Animal = (function () {
    function Animal() {
        this.move();
    }
    Animal.prototype.move = function (distanceInMeters) {
        if (distanceInMeters === void 0) { distanceInMeters = 0; }
        console.log(distanceInMeters);
    };
    return Animal;
}());
var Snake = (function (_super) {
    __extends(Snake, _super);
    function Snake() {
        _super.call(this);
        this.variable = 1;
    }
    Snake.prototype.move = function (distanceInMeters) {
        if (distanceInMeters === void 0) { distanceInMeters = 5; }
        console.log(this.variable);
        _super.prototype.move.call(this, distanceInMeters);
    };
    return Snake;
}(Animal));

[yortus]

I believe the compiler is working correctly here.

In ES6 classes, you cannot reference this before calling super (MDN ref, StackOverflow ref). this will not be bound until after the super() call. You can try it for yourself using node.js or the browser console: referencing this in the constructor before calling super() will throw something like "ReferenceError: this is undefined".

The equivalent ES3/5 downlevelled code has to do things in that order too, otherwise it would not be runtime-equivalent code.

[SimonCsoma]

I expect this to complie like this to ES6:

class Animal {
    constructor() { 
        this.move();
    }
    move(distanceInMeters = 0) {
        console.log(distanceInMeters);
    }
}

class _Animal {
    constructor() {
        this.variable = 1
        this.move();
    }
    move(distanceInMeters = 0) {
        console.log(distanceInMeters);
    }
}


class Snake extends _Animal{
    constructor() { 
        super();

    }
    move(distanceInMeters = 5) {
        console.log(this.variable);
        super.move(distanceInMeters);
    }
}

new Snake();

I know this does not make allot of sense on the first view, but since typescript supports public variable I expect it to always be available and this is for now the only way i can produce the expected behaviour.

In my opinion if this cannot be supported it should be removed from the language or a fix shuold be added like this.

[RyanCavanaugh]

Duplicate #1617.