I'm trying to add extra methods to class, and these extra methods should use the super methods.
If I add them in the model definition, it works.
class A {
doSomething() {
console.log('logSomething');
}
}
class B extends A {
doSomething() {
super.doSomething();
console.log('logSomethingElse');
}
}
If I try to add the extra method to B.prototype, I'll get SyntaxError: 'super' keyword unexpected here.
class A {
doSomething() {
console.log('logSomething');
}
}
class B extends A {
}
B.prototype.doSomething = function doSomething() {
super.doSomething();
console.log('logSomethingElse');
}
It is quite clear, why I get this error. This is a function and not a class method.
Let's try to define the method as a class method, and copy it to the original B class:
class A {
doSomething() {
console.log('logSomething');
}
}
class B extends A {}
class X {
doSomething() {
super.doSomething();
console.log('2 logSomethingElse');
}
}
B.prototype.doSomething = X.prototype.doSomething;
In this case I'll get TypeError: (intermediate value).doSomething is not a function.
Is there any way to define methods (that refer to super) outside from the original class definition, and add these methods later to the original class?
super refers to ancestor of a class where the method was defined, it isn't dynamic. As Babel output illustrates this, super is hard-coded to Object.getPrototypeOf(X.prototype), and thus orphan class like this one doesn't make sense because it doesn't have super:
class X {
doSomething() {
super.doSomething();
...
}
}
But super can be substituted with dynamic counterpart:
doSomething() {
const dynamicSuper = Object.getPrototypeOf(this.constructor.prototype);
// or
// const dynamicSuper = Object.getPrototypeOf(Object.getPrototypeOf(this));
dynamicSuper.doSomething();
...
}
class B extends A {}
B.prototype.doSomething = doSomething;
In this case it will refer to ancestor class of class instance where doSomething was assigned as prototype method.
While I think this could be assumed as anti-pattern, you shouldn't use super outside from a class.
You can achieve that using Object Literals.
Refer to Object.setPrototypeOf
const A = {
sayHello() {
console.log("I am A");
},
Factory() {
return Object.create(this);
}
}
const B = {
sayHello() {
super.sayHello();
}
}
Object.setPrototypeOf(B, A);
const c = B.Factory();
c.sayHello();
If you don't make class X inherit from class B or A, the only way to call the method is A.prototype.doSomething() or more generally A.prototype.doSomething.call(this_substitute, ...args).
Related
Let's take this exemple :
class A {
attrA = 3;
}
class B {
constructor() {
this.a = new A;
}
attrB = this.a.attrA;
methB() { console.log(this.a.attrA);}
}
const test = new B;
console.log(test.attrB);
test.methB();
I can access the class A attribute through method of class B, but I can't use it with attribute of class B, I have an "undefined" error.
The only way to do this is :
class A {
attrA = 3;
}
class B {
constructor() {
this.a = new A;
this.z = this.a.attrA
}
methB() { console.log(this.a.attrA);}
}
const test = new B;
console.log(test.z);
test.methB();
Why I need to put the attribute in the constructor and not the method?
Thanks!
You can think of class fields as being hoisted, meaning this:
class Example {
constructor() {/*...*/}
property = "value";
}
Is "actually" this:
class Example {
property = "value";
constructor() {/*...*/}
}
Or (similar to) this:
class Example {
constructor() {
this.property = "value";
/*...*/
}
}
Further, identifiers are resolved upon access. So by the time you execute test.methB(), test.a has been initialized, allowing the method to correctly resolve this.a.attrA. It works the same as this code:
let variable = null;
const logVariable = () => console.log(variable); // You may think this logs null, but ...
variable = "value";
logVariable(); // ... it actually logs `variable`'s value at the time of calling.
As you have observed, mixing property initializations from within the constructor and using field initializers may be confusing. Therefore, a more intuitive way to write your code would be:
class A {
attrA = 3;
}
class B {
a = new A;
attrB = this.a.attrA;
constructor() {}
methB() {
console.log(this.a.attrA);
}
}
const test = new B;
console.log(test.attrB); // Works now!
test.methB();
Personal recommendations:
Declare all instance and class fields.
Prefer field initializers.
Only reassign/initialize fields in the constructor for non-default values.
You may want to read on for more details for a better technical understanding.
Class syntax
Classes are just uncallable constructor functions:
class Example {}
console.log(typeof Example); // "function"
Example(); // Error: cannot be invoked without `new`
This is a quirk of the ES6 class syntax. Technically, class constructors are still "callable", but only internally (which happens when using new). Otherwise constructors would serve no purpose.
Another aspect of the class syntax is the ability to call super() in the constructor. This only comes into play when a class inherits, but that comes with yet its own quirks:
You cannot use this before calling super:
class A {};
class B extends A {
constructor() {
this; // Error: must call super before accessing this
super();
}
}
new B();
Reason being, before calling super no object has been created, despite having used new and being in the constructor.
The actual object creation happens at the base class, which is in the most nested call to super. Only after calling super has an object been created, allowing the use of this in the respective constructor.
Class fields
With the addition of instance fields in ES13, constructors became even more complicated: Initialization of those fields happens immediately after the object creation or the call to super, meaning before the statements that follow.
class A /*no heritage*/ {
property = "a";
constructor() {
// Initializing instance fields
// ... (your code)
}
}
class B extends A {
property = "b";
constructor() {
super();
// Initializing instance fields
// ... (your code)
}
}
Further, those property "assignments" are actually no assignments but definitions:
class Setter {
set property(value) {
console.log("Value:", value);
}
}
class A extends Setter {
property = "from A";
}
class B extends Setter {
constructor() {
super();
// Works effectively like class A:
Object.defineProperty(this, "property", { value: "from B" });
}
}
class C extends Setter {
constructor() {
super();
this.property = "from C";
}
}
new A(); // No output, expected: "Value: from A"
new B(); // No output, expected: "Value: from B"
new C(); // Output: "Value: from C"
Variables
Identifiers are only resolved upon access, allowing this unintuitive code:
const logNumber = () => console.log(number); // References `number` before its declaration, but ...
const number = 0;
logNumber(); // ... it is resolved here, making it valid.
Also, identifiers are looked up from the nearest to the farthest lexical environment. This allows variable shadowing:
const variable = "initial-value";
{
const variable = "other-value"; // Does NOT reassign, but *shadows* outer `variable`.
console.log(variable);
}
console.log(variable);
I have class A which is parent of class B
class A {
constructor(a){
this.a=a;
}
par(){
console.log("para");
}
}
class B extends A {
constructor(a) {
super(a)
this.a = "child";
}
par() {
super.par();
console.log("child");
}
}
When I use this code, it works fine.
But when I explicitly define the par function on B using this code:
B.prototype.par = function() {
super.par();
}
I get the error
Uncaught SyntaxError: 'super' keyword unexpected here
Whether we create a function in class definition or in prototype object of function('class'), it should be the same thing.
What am I doing wrong here?
'super' is simply a syntactic sugar introduced in ES2015 along with class syntax.
It can only be used within functions of 'class' (constructor and methods) that extends another class.
class A {
constructor(){}
par(){ console.log('para') }
}
class B extends A {
constructor(){
super()
}
}
Is equivalent to:
function A(){}
A.prototype.par = function(){console.log('para')}
var B = (function(parent){
var _super = parent;
function B(){
_super.call(this); // calls parent's constructor
}
B.prototype = Object.create(_super.prototype); // Inherits parent's methods.
B.prototype.par = function(){ // override parent's par.
_super.prototype.par.call(this); // child still has access to parent's par method thanks to closure :)
console.log('child');
}
return B;
})(A);
var b = new B();
b.par()
You cannot do:
function(){
super // super is not defined...
}
I couldn't find the solution to this issue. I am banging my head against a wall entire day.
Assume we have the following hierarchy of classes:
class A {
async name() { ... }
}
class B extends A {
async age() { ... }
}
class C extends A {
async group() { ... }
}
class D extends B {
constructor() {
super(...)
this.C = new C(...);
}
async group() { return await this.C.group(); }
}
Because JavaScript doesn't support multiple inheritance, I have an issue. I would like to combine prototypes of objects in a way that I can invoke function group() without the need for writing repetitive code like above. It is absolute must to be able to invoke group() directly on the object D outside of the class, even though it is the an instance of class C.
How can I combine their prototypes? I've tried things like
class A {
appendPrototype(obj) {
if (obj instanceof A)
for (let k in obj.__proto__) {
this.__proto__[k] = obj.__proto__[k];
}
}
}
.
.
. // Same two classes, B and C
class D extends B {
constructor(...) {
super(...);
const c = new C(...);
this.appendPrototype(c);
}
}
I can't invoke the function group() an instance of the class D. Why is that? How can I achieve it, without writing repetitive code like in the first example?
What you can do, it is to assign to the prototype of the class D, the method of the class C.
In this way all the instances of D will have the group method which is the same method defined within the class C, and you can do the same for all the other child classes.
class A {
async name() {
console.log('A - name');
}
}
class B extends A {
async age() {
console.log('B - age');
}
}
class C extends A {
async group() {
console.log('C - group');
}
}
class D extends B {
constructor() {
super();
}
}
D.prototype.group = C.prototype.group;
const d = new D();
d.group();
Anyway, in your case, I would go for composition (like you did with C within D) and then call the methods like:
d.c.group()
which allows you to avoid the appendPrototype logic.
Is it possible to extend a class in ES6 without calling the super method to invoke the parent class?
EDIT: The question might be misleading. Is it the standard that we have to call super() or am I missing something?
For example:
class Character {
constructor(){
console.log('invoke character');
}
}
class Hero extends Character{
constructor(){
super(); // exception thrown here when not called
console.log('invoke hero');
}
}
var hero = new Hero();
When I'm not calling super() on the derived class I'm getting a scope problem -> this is not defined
I'm running this with iojs --harmony in v2.3.0
The rules for ES2015 (ES6) classes basically come down to:
In a child class constructor, this cannot be used until super is called.
ES6 class constructors MUST call super if they are subclasses, or they must explicitly return some object to take the place of the one that was not initialized.
This comes down to two important sections of the ES2015 spec.
Section 8.1.1.3.4 defines the logic to decide what this is in the function. The important part for classes is that it is possible for this be in an "uninitialized" state, and when in this state, attempting to use this will throw an exception.
Section 9.2.2, [[Construct]], which defines the behavior of functions called via new or super. When calling a base class constructor, this is initialized at step #8 of [[Construct]], but for all other cases, this is uninitialized. At the end of construction, GetThisBinding is called, so if super has not been called yet (thus initializing this), or an explicit replacement object was not returned, the final line of the constructor call will throw an exception.
The new ES6 class syntax is only an other notation for "old" ES5 "classes" with prototypes. Therefore you cannot instantiate a specific class without setting its prototype (the base class).
Thats like putting cheese on your sandwich without making it. Also you cannot put cheese before making the sandwich, so...
...using this keyword before calling the super class with super() is not allowed, too.
// valid: Add cheese after making the sandwich
class CheeseSandwich extend Sandwich {
constructor() {
super();
this.supplement = "Cheese";
}
}
// invalid: Add cheese before making sandwich
class CheeseSandwich extend Sandwich {
constructor() {
this.supplement = "Cheese";
super();
}
}
// invalid: Add cheese without making sandwich
class CheeseSandwich extend Sandwich {
constructor() {
this.supplement = "Cheese";
}
}
If you don’t specify a constructor for a base class, the following definition is used:
constructor() {}
For derived classes, the following default constructor is used:
constructor(...args) {
super(...args);
}
EDIT: Found this on developer.mozilla.org:
When used in a constructor, the super keyword appears alone and must be used before the this keyword can be used.
Source
There have been multiple answers and comments stating that super MUST be the first line inside constructor. That is simply wrong. #loganfsmyth answer has the required references of the requirements, but it boil down to:
Inheriting (extends) constructor must call super before using this and before returning even if this isn't used
See fragment below (works in Chrome...) to see why it might make sense to have statements (without using this) before calling super.
'use strict';
var id = 1;
function idgen() {
return 'ID:' + id++;
}
class Base {
constructor(id) {
this.id = id;
}
toString() { return JSON.stringify(this); }
}
class Derived1 extends Base {
constructor() {
var anID = idgen() + ':Derived1';
super(anID);
this.derivedProp = this.baseProp * 2;
}
}
alert(new Derived1());
You can omit super() in your subclass, if you omit the constructor altogether in your subclass. A 'hidden' default constructor will be included automatically in your subclass. However, if you do include the constructor in your subclass, super() must be called in that constructor.
class A{
constructor(){
this.name = 'hello';
}
}
class B extends A{
constructor(){
// console.log(this.name); // ReferenceError
super();
console.log(this.name);
}
}
class C extends B{} // see? no super(). no constructor()
var x = new B; // hello
var y = new C; // hello
Read this for more information.
The answer by justyourimage is the easiest way, but his example is a little bloated. Here's the generic version:
class Base {
constructor(){
return this._constructor(...arguments);
}
_constructor(){
// just use this as the constructor, no super() restrictions
}
}
class Ext extends Base {
_constructor(){ // _constructor is automatically called, like the real constructor
this.is = "easy"; // no need to call super();
}
}
Don't extend the real constructor(), just use the fake _constructor() for the instantiation logic.
Note, this solution makes debugging annoying because you have to step into an extra method for every instantiation.
Just registered to post this solution since the answers here don't satisfy me the least since there is actually a simple way around this. Adjust your class-creation pattern to overwrite your logic in a sub-method while using only the super constructor and forward the constructors arguments to it.
As in you do not create an constructor in your subclasses per se but only reference to an method that is overridden in the respective subclass.
That means you set yourself free from the constructor functionality enforced upon you and refrain to a regular method - that can be overridden and doesn't enforce super() upon you letting yourself the choice if, where and how you want to call super (fully optional) e.g.:
super.ObjectConstructor(...)
class Observable {
constructor() {
return this.ObjectConstructor(arguments);
}
ObjectConstructor(defaultValue, options) {
this.obj = { type: "Observable" };
console.log("Observable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
class ArrayObservable extends Observable {
ObjectConstructor(defaultValue, options, someMoreOptions) {
this.obj = { type: "ArrayObservable" };
console.log("ArrayObservable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
class DomainObservable extends ArrayObservable {
ObjectConstructor(defaultValue, domainName, options, dependent1, dependent2) {
this.obj = super.ObjectConstructor(defaultValue, options);
console.log("DomainObservable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
var myBasicObservable = new Observable("Basic Value", "Basic Options");
var myArrayObservable = new ArrayObservable("Array Value", "Array Options", "Some More Array Options");
var myDomainObservable = new DomainObservable("Domain Value", "Domain Name", "Domain Options", "Dependency A", "Depenency B");
cheers!
#Bergi mentioned new.target.prototype, but I was looking for a concrete example proving that you can access this (or better, the reference to the object the client code is creating with new, see below) without having to call super() at all.
Talk is cheap, show me the code... So here is an example:
class A { // Parent
constructor() {
this.a = 123;
}
parentMethod() {
console.log("parentMethod()");
}
}
class B extends A { // Child
constructor() {
var obj = Object.create(new.target.prototype)
// You can interact with obj, which is effectively your `this` here, before returning
// it to the caller.
return obj;
}
childMethod(obj) {
console.log('childMethod()');
console.log('this === obj ?', this === obj)
console.log('obj instanceof A ?', obj instanceof A);
console.log('obj instanceof B ?', obj instanceof B);
}
}
b = new B()
b.parentMethod()
b.childMethod(b)
Which will output:
parentMethod()
childMethod()
this === obj ? true
obj instanceof A ? true
obj instanceof B ? true
So you can see that we are effectively creating an object of type B (the child class) which is also an object of type A (its parent class) and within the childMethod() of child B we have this pointing to the object obj which we created in B's constructor with Object.create(new.target.prototype).
And all this without caring about super at all.
This leverages the fact that in JS a constructor can return a completely different object when the client code constructs a new instance with new.
Hope this helps someone.
Try:
class Character {
constructor(){
if(Object.getPrototypeOf(this) === Character.prototype){
console.log('invoke character');
}
}
}
class Hero extends Character{
constructor(){
super(); // throws exception when not called
console.log('invoke hero');
}
}
var hero = new Hero();
console.log('now let\'s invoke Character');
var char = new Character();
Demo
I would recommend to use OODK-JS if you intend to develop following OOP concepts.
OODK(function($, _){
var Character = $.class(function ($, µ, _){
$.public(function __initialize(){
$.log('invoke character');
});
});
var Hero = $.extends(Character).class(function ($, µ, _){
$.public(function __initialize(){
$.super.__initialize();
$.log('invoke hero');
});
});
var hero = $.new(Hero);
});
Simple solution: I think its clear no need for explanation.
class ParentClass() {
constructor(skipConstructor = false) { // default value is false
if(skipConstructor) return;
// code here only gets executed when 'super()' is called with false
}
}
class SubClass extends ParentClass {
constructor() {
super(true) // true for skipping ParentClass's constructor.
// code
}
}
Is it possible to extend a class in ES6 without calling the super method to invoke the parent class?
EDIT: The question might be misleading. Is it the standard that we have to call super() or am I missing something?
For example:
class Character {
constructor(){
console.log('invoke character');
}
}
class Hero extends Character{
constructor(){
super(); // exception thrown here when not called
console.log('invoke hero');
}
}
var hero = new Hero();
When I'm not calling super() on the derived class I'm getting a scope problem -> this is not defined
I'm running this with iojs --harmony in v2.3.0
The rules for ES2015 (ES6) classes basically come down to:
In a child class constructor, this cannot be used until super is called.
ES6 class constructors MUST call super if they are subclasses, or they must explicitly return some object to take the place of the one that was not initialized.
This comes down to two important sections of the ES2015 spec.
Section 8.1.1.3.4 defines the logic to decide what this is in the function. The important part for classes is that it is possible for this be in an "uninitialized" state, and when in this state, attempting to use this will throw an exception.
Section 9.2.2, [[Construct]], which defines the behavior of functions called via new or super. When calling a base class constructor, this is initialized at step #8 of [[Construct]], but for all other cases, this is uninitialized. At the end of construction, GetThisBinding is called, so if super has not been called yet (thus initializing this), or an explicit replacement object was not returned, the final line of the constructor call will throw an exception.
The new ES6 class syntax is only an other notation for "old" ES5 "classes" with prototypes. Therefore you cannot instantiate a specific class without setting its prototype (the base class).
Thats like putting cheese on your sandwich without making it. Also you cannot put cheese before making the sandwich, so...
...using this keyword before calling the super class with super() is not allowed, too.
// valid: Add cheese after making the sandwich
class CheeseSandwich extend Sandwich {
constructor() {
super();
this.supplement = "Cheese";
}
}
// invalid: Add cheese before making sandwich
class CheeseSandwich extend Sandwich {
constructor() {
this.supplement = "Cheese";
super();
}
}
// invalid: Add cheese without making sandwich
class CheeseSandwich extend Sandwich {
constructor() {
this.supplement = "Cheese";
}
}
If you don’t specify a constructor for a base class, the following definition is used:
constructor() {}
For derived classes, the following default constructor is used:
constructor(...args) {
super(...args);
}
EDIT: Found this on developer.mozilla.org:
When used in a constructor, the super keyword appears alone and must be used before the this keyword can be used.
Source
There have been multiple answers and comments stating that super MUST be the first line inside constructor. That is simply wrong. #loganfsmyth answer has the required references of the requirements, but it boil down to:
Inheriting (extends) constructor must call super before using this and before returning even if this isn't used
See fragment below (works in Chrome...) to see why it might make sense to have statements (without using this) before calling super.
'use strict';
var id = 1;
function idgen() {
return 'ID:' + id++;
}
class Base {
constructor(id) {
this.id = id;
}
toString() { return JSON.stringify(this); }
}
class Derived1 extends Base {
constructor() {
var anID = idgen() + ':Derived1';
super(anID);
this.derivedProp = this.baseProp * 2;
}
}
alert(new Derived1());
You can omit super() in your subclass, if you omit the constructor altogether in your subclass. A 'hidden' default constructor will be included automatically in your subclass. However, if you do include the constructor in your subclass, super() must be called in that constructor.
class A{
constructor(){
this.name = 'hello';
}
}
class B extends A{
constructor(){
// console.log(this.name); // ReferenceError
super();
console.log(this.name);
}
}
class C extends B{} // see? no super(). no constructor()
var x = new B; // hello
var y = new C; // hello
Read this for more information.
The answer by justyourimage is the easiest way, but his example is a little bloated. Here's the generic version:
class Base {
constructor(){
return this._constructor(...arguments);
}
_constructor(){
// just use this as the constructor, no super() restrictions
}
}
class Ext extends Base {
_constructor(){ // _constructor is automatically called, like the real constructor
this.is = "easy"; // no need to call super();
}
}
Don't extend the real constructor(), just use the fake _constructor() for the instantiation logic.
Note, this solution makes debugging annoying because you have to step into an extra method for every instantiation.
Just registered to post this solution since the answers here don't satisfy me the least since there is actually a simple way around this. Adjust your class-creation pattern to overwrite your logic in a sub-method while using only the super constructor and forward the constructors arguments to it.
As in you do not create an constructor in your subclasses per se but only reference to an method that is overridden in the respective subclass.
That means you set yourself free from the constructor functionality enforced upon you and refrain to a regular method - that can be overridden and doesn't enforce super() upon you letting yourself the choice if, where and how you want to call super (fully optional) e.g.:
super.ObjectConstructor(...)
class Observable {
constructor() {
return this.ObjectConstructor(arguments);
}
ObjectConstructor(defaultValue, options) {
this.obj = { type: "Observable" };
console.log("Observable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
class ArrayObservable extends Observable {
ObjectConstructor(defaultValue, options, someMoreOptions) {
this.obj = { type: "ArrayObservable" };
console.log("ArrayObservable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
class DomainObservable extends ArrayObservable {
ObjectConstructor(defaultValue, domainName, options, dependent1, dependent2) {
this.obj = super.ObjectConstructor(defaultValue, options);
console.log("DomainObservable ObjectConstructor called with arguments: ", arguments);
console.log("obj is:", this.obj);
return this.obj;
}
}
var myBasicObservable = new Observable("Basic Value", "Basic Options");
var myArrayObservable = new ArrayObservable("Array Value", "Array Options", "Some More Array Options");
var myDomainObservable = new DomainObservable("Domain Value", "Domain Name", "Domain Options", "Dependency A", "Depenency B");
cheers!
#Bergi mentioned new.target.prototype, but I was looking for a concrete example proving that you can access this (or better, the reference to the object the client code is creating with new, see below) without having to call super() at all.
Talk is cheap, show me the code... So here is an example:
class A { // Parent
constructor() {
this.a = 123;
}
parentMethod() {
console.log("parentMethod()");
}
}
class B extends A { // Child
constructor() {
var obj = Object.create(new.target.prototype)
// You can interact with obj, which is effectively your `this` here, before returning
// it to the caller.
return obj;
}
childMethod(obj) {
console.log('childMethod()');
console.log('this === obj ?', this === obj)
console.log('obj instanceof A ?', obj instanceof A);
console.log('obj instanceof B ?', obj instanceof B);
}
}
b = new B()
b.parentMethod()
b.childMethod(b)
Which will output:
parentMethod()
childMethod()
this === obj ? true
obj instanceof A ? true
obj instanceof B ? true
So you can see that we are effectively creating an object of type B (the child class) which is also an object of type A (its parent class) and within the childMethod() of child B we have this pointing to the object obj which we created in B's constructor with Object.create(new.target.prototype).
And all this without caring about super at all.
This leverages the fact that in JS a constructor can return a completely different object when the client code constructs a new instance with new.
Hope this helps someone.
Try:
class Character {
constructor(){
if(Object.getPrototypeOf(this) === Character.prototype){
console.log('invoke character');
}
}
}
class Hero extends Character{
constructor(){
super(); // throws exception when not called
console.log('invoke hero');
}
}
var hero = new Hero();
console.log('now let\'s invoke Character');
var char = new Character();
Demo
I would recommend to use OODK-JS if you intend to develop following OOP concepts.
OODK(function($, _){
var Character = $.class(function ($, µ, _){
$.public(function __initialize(){
$.log('invoke character');
});
});
var Hero = $.extends(Character).class(function ($, µ, _){
$.public(function __initialize(){
$.super.__initialize();
$.log('invoke hero');
});
});
var hero = $.new(Hero);
});
Simple solution: I think its clear no need for explanation.
class ParentClass() {
constructor(skipConstructor = false) { // default value is false
if(skipConstructor) return;
// code here only gets executed when 'super()' is called with false
}
}
class SubClass extends ParentClass {
constructor() {
super(true) // true for skipping ParentClass's constructor.
// code
}
}