Accessing member of 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:
- The first log outputs
super.name
, this refers to the prototype chain of the objecttom
, not the objecttom
self. Because we have added a name property on theAnimal.prototype
, horseee will be outputted. - The second log outputs
this.name
, thethis
keyword refers to the the tom object itself. - 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 syntaxsuper.move(45);
. Using thesuper
keyword in this context will look for amove
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.
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);
You don't need to manually assign the name to a public variable. Using
public name
in the constructor definition does this for you.You don't need to call
super(name)
from the specialised classes.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.