Accessing member of base class

See the inheritance example from the playground on the TypeScript site:

class Animal {
  public name;
  constructor(name) {
    this.name = name;
  }
  move(meters) {
    alert(this.name + " moved " + meters + "m.");
  }
}

class Snake extends Animal {
  constructor(name) {
    super(name);
  }
  move() {
    alert("Slithering...");
    super.move(5);
  }
}

class Horse extends Animal {
  constructor(name) {
    super(name);
  }
  move() {
    alert(super.name + " is Galloping...");
    super.move(45);
  }
}

var sam = new Snake("Sammy the Python");
var tom: Animal = new Horse("Tommy the Palomino");

sam.move();
tom.move(34);

I have changed one line of code: the alert in Horse.move(). There I want to access super.name, but that returns just undefined. IntelliSense is suggesting that I can use it and TypeScript compiles fine, but it does not work.

Any ideas?

Working example. Notes below.

class Animal {
    constructor(public name) {
    }

    move(meters) {
        alert(this.name + " moved " + meters + "m.");
    }
}

class Snake extends Animal {
    move() {
        alert(this.name + " is Slithering...");
        super.move(5);
    }
}

class Horse extends Animal {
    move() {
        alert(this.name + " is Galloping...");
        super.move(45);
    }
}

var sam = new Snake("Sammy the Python");
var tom: Animal = new Horse("Tommy the Palomino");

sam.move();
tom.move(34);

1. You don't need to manually assign the name to a public variable. Using public name in the constructor definition does this for you.

2. You don't need to call super(name) from the specialised classes.

3. Using this.name works.

Notes on use of super.

This is covered in more detail in section 4.9.2 of the language specification.

The behaviour of the classes inheriting from Animal is not dissimilar to the behaviour in other languages. You need to specify the super keyword in order to avoid confusion between a specialised function and the base class function. For example, if you called move() or this.move() you would be dealing with the specialised Snake or Horse function, so using super.move() explicitly calls the base class function.

There is no confusion of properties, as they are the properties of the instance. There is no difference between super.name and this.name - there is simply this.name. Otherwise you could create a Horse that had different names depending on whether you were in the specialized class or the base class.

You are incorrectly using the super and this keyword. Here is an example of how they work:

class Animal {
    public name: string;
    constructor(name: string) { 
        this.name = name;
    }
    move(meters: number) {
        console.log(this.name + " moved " + meters + "m.");
    }
}

class Horse extends Animal {
    move() {
        console.log(super.name + " is Galloping...");
        console.log(this.name + " is Galloping...");
        super.move(45);
    }
}

var tom: Animal = new Horse("Tommy the Palomino");

Animal.prototype.name = 'horseee'; 

tom.move(34);
// Outputs:

// horseee is Galloping...
// Tommy the Palomino is Galloping...
// Tommy the Palomino moved 45m.

Explanation:

1. The first log outputs super.name, this refers to the prototype chain of the object tom, not the object tom self. Because we have added a name property on the Animal.prototype, horseee will be outputted.
2. The second log outputs this.name, the this keyword refers to the the tom object itself.
3. The third log is logged using the move method of the Animal base class. This method is called from Horse class move method with the syntax super.move(45);. Using the super keyword in this context will look for a move method on the prototype chain which is found on the Animal prototype.

Remember TS still uses prototypes under the hood and the class and extends keywords are just syntactic sugar over prototypical inheritance.