Call an overridden method from super class in typescript
The key is calling the parent's method using super.methodName();
class A {
// A protected method
protected doStuff()
{
alert("Called from A");
}
// Expose the protected method as a public function
public callDoStuff()
{
this.doStuff();
}
}
class B extends A {
// Override the protected method
protected doStuff()
{
// If we want we can still explicitly call the initial method
super.doStuff();
alert("Called from B");
}
}
var a = new A();
a.callDoStuff(); // Will only alert "Called from A"
var b = new B()
b.callDoStuff(); // Will alert "Called from A" then "Called from B"
Try it here
The order of execution is:
A
's constructorB
's constructor
The assignment occurs in B
's constructor after A
's constructor—_super
—has been called:
function B() {
_super.apply(this, arguments); // MyvirtualMethod called in here
this.testString = "Test String"; // testString assigned here
}
So the following happens:
var b = new B(); // undefined
b.MyvirtualMethod(); // "Test String"
You will need to change your code to deal with this. For example, by calling this.MyvirtualMethod()
in B
's constructor, by creating a factory method to create the object and then execute the function, or by passing the string into A
's constructor and working that out somehow... there's lots of possibilities.
If you want a super class to call a function from a subclass, the cleanest way is to define an abstract pattern, in this manner you explicitly know the method exists somewhere and must be overridden by a subclass.
This is as an example, normally you do not call a sub method within the constructor as the sub instance is not initialized yet… (reason why you have an "undefined" in your question's example)
abstract class A {
// The abstract method the subclass will have to call
protected abstract doStuff():void;
constructor(){
alert("Super class A constructed, calling now 'doStuff'")
this.doStuff();
}
}
class B extends A{
// Define here the abstract method
protected doStuff()
{
alert("Submethod called");
}
}
var b = new B();
Test it Here
And if like @Max you really want to avoid implementing the abstract method everywhere, just get rid of it. I don't recommend this approach because you might forget you are overriding the method.
abstract class A {
constructor() {
alert("Super class A constructed, calling now 'doStuff'")
this.doStuff();
}
// The fallback method the subclass will call if not overridden
protected doStuff(): void {
alert("Default doStuff");
};
}
class B extends A {
// Override doStuff()
protected doStuff() {
alert("Submethod called");
}
}
class C extends A {
// No doStuff() overriding, fallback on A.doStuff()
}
var b = new B();
var c = new C();
Try it Here