On top of being irritating and requiring modifications to every child class whenever the parent class is updated...
On top of being irritating and requiring modifications to every child class whenever the parent class is updated...
Let's explore the code provided below:
class A {
protected total: number;
constructor(protected num1: number, protected num2: number) {
this.total = this.num1 + this.num2;
}
}
class B extends A {
constructor(num1: number, num2: number) {
super(num1, num2);
}
}
In the constructor of class B, the call to super invokes the constructor of class A. When we examine the compiled javascript code:
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 A = (function () {
function A(num1, num2) {
this.num1 = num1;
this.num2 = num2;
this.total = this.num1 + this.num2;
}
return A;
}());
var B = (function (_super) {
__extends(B, _super);
function B(num1, num2) {
_super.call(this, num1, num2);
}
return B;
}(A));
The reason behind explicitly calling super is to ensure that the constructor of class A is executed and members like num1, num2, and total are initialized in instances of class B.
You may wonder why the compiler doesn't automatically call super, and there are a couple of reasons for that:
(1) There may be scenarios where actions need to be taken before invoking super, as shown below:
class A {
protected sum: number;
constructor(protected x: number, protected y: number) {
this.sum = this.x + this.y;
}
}
class B extends A {
constructor(x: number, y: number) {
if (x % 2 === 0) {
super(x, y);
} else {
super(x + 1, y);
}
}
}
It's essential to call super prior to accessing this in the constructor of class B.
(2) Explicitly calling super makes it clear that the parent constructor is being invoked, preventing any unexpected behavior.
This requirement only applies to constructors; methods within a class do not need to call their superclass, but you have the option to do so if you wish to inherit functionality from the parent method.
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