I don't want my parent class to be too long so I separate some methods from it and create child class.
However I don't want to use child class as a instance I want it to be used only by parent class.
class Parent {
parentMethod() {
this.foo(); // execute from parent
}
}
class Child extends Parent {
foo() {
console.log('foo!');
}
}
const parent = new Parent();
parent.parentMethod(); // execute parent method which execute child method
this cause:
Uncaught TypeError: this.foo is not a function
I don't want my parent class to be too long so I separate some methods from it
Ok.
So I create child class, however I don't want to use child class as a instance.
No, subclassing is the wrong approach here. Especially if you don't want to instantiate the subclass, it's not even a solution to your problem.
To separate units of code, factor them out into separate functions. Those don't need to be linked to the caller through inheritance, and don't need to be methods of a class at all. Just write
class MyClass {
myMethod() {
foo();
}
}
function foo() {
console.log('foo!');
}
const instance = new MyClass();
instance.myMethod();
Or compose your object of multiple smaller helpers:
class Outer {
constructor() {
this.helper = new Inner();
}
myMethod() {
this.helper.foo();
}
}
class Inner {
foo() {
console.log('foo!');
}
}
const instance = new Outer();
instance.myMethod();
If you want to use the Parent class in the subclass, Child class, then you need to do the following:
class Parent {
foo() {
console.log('foo!');
}
}
class Child extends Parent {
constructor() {
super();
}
}
let c = new Child(); //instantiate Child class
c.foo(); // here you are calling super.foo(); which is the Parent classes method for foo.
//foo!
The super keyword is used to access and call functions on an object's
parent.
How to use super
Or, alternatively, if you'd rather create a method on the child class that wraps the parent method of foo, rather than accessing it by instantiating the parent class via calling super in the child constructor:
class Parent {
foo() {
console.log('foo!');
}
}
class Child extends Parent {
method() {
this.foo();
}
}
let c = new Child();
c.method();
Related
This is my first question on Stackoverflow so please dont run over me like a bulldozer if i did something wrong :)
I need to know if its possible in JavaScript classes to know, if the child has provided a constructor.
E.g.
class Parent {
constructor() {
console.log('Child has constructor:', /* Magic here */)
}
}
class Child extends Parent {}
new Child()
Expected output: Child has constructor: false
Vs:
class Parent {
constructor() {
console.log('Child has constructor:', /* Magic here */)
}
}
class Child extends Parent {
constructor() {
super()
}
}
new Child()
Expected output: Child has constructor: true
Background: I would like to have a class that behaves differently when it was extended than if it was used directly.
Since Childs should provide the Parent different informations than if it was used directly.
You could add one parameter to the Parent constructor that is false by default and then when you call the super inside the Child class you pass true for this parameter.
class Parent {
constructor(called = false) {
console.log('Child has constructor:', called)
}
}
class Child extends Parent {
constructor() {
super(true)
}
}
class ChildTwo extends Parent {}
new Child()
new ChildTwo()
No, it's not possible.
The class syntax in JavaScript is just a syntax suger of a normal function.
Take this ES6 example:
class Parent {
constructor(value){
this.example = value;
}
parentMethod(){
console.log('parent:', this.example);
}
}
class Child extends Parent {
childMethod(){
console.log('children:', this.example);
}
}
const parent = new Parent('Hello');
const child = new Child('World');
parent.parentMethod();
child.childMethod();
console.log(parent.constructor);
console.log(child.constructor);
As you can see, even if you don't explicitly define a constructor, a class will always have a constructor.
The above could roughly translate into the below ES5 code which does not yet support class syntax:
function Parent(value){
this.example = value;
}
Object.defineProperties(Parent.prototype, {
parentMethod: {
writable: true,
enumerable: false,
configurable: true,
value: function(){
console.log('parent:', this.example);
}
}
});
function Child(value){
Parent.call(this, value);
}
Child.prototype = Object.create(Parent.prototype);
Child.prototype.constructor = Child;
Child.prototype = Object.defineProperties(Child.prototype, {
childMethod: {
writable: true,
enumerable: false,
configurable: true,
value: function(){
console.log('child:', this.example);
}
}
});
var parent = new Parent('Hello');
var child = new Child('World');
parent.parentMethod();
child.childMethod();
console.log(parent.constructor);
console.log(child.constructor);
As you can see:
A class is merely just a function.
The .constructor is always assigned.
Hence, there is no way for you to check if child class has a constructor because constructor is always there.
Even if you can do it (which you can't), the Parent will not know beforehand what class will extend it, so it will not know whether or not a future child will have a constructor or not.
I need to know if its possible in JavaScript classes to know, if the child has provided a constructor.
There are no classes without a constructor. Some classes might have an implicit constructor (no constructor in the class syntax), but you cannot - and should not - detect that.
I would like to have a class that behaves differently when it was extended than if it was used directly
You can distinguish a new Child from a new Parent call using new.target == Parent.
Childs should provide the Parent different information than if it was used directly.
That's a bad idea. Your parent constructor should have an interface that doesn't care about who is using it. I would suggest to either
never use your parent class directly, but extend it yourself and provide that class for default usage. It can handle the different information.
make the parent class accept different kinds of information, i.e. overload its interface. Distinguish what information you were given, and behave accordingly. However, it should not matter whether the information was given directly by a user or produced by a child class.
I've recently started to learn about Classes in javascript and while reading some really interesting stuff I thought of trying some of my own ideas.
If you have a parent class of Parent in which you have a method of logSomething```` and a child class ofChild, with which you doclass Child extends Parent, how can you then execute the inherited method from the parent class,logSomething```, inside of the child class?
If you define a method inside of the Child class and add this.logSomething() to that method, whenever the method from the child class is called, the inherited logSomething function will indeed run, but apart from that I haven't found any way of executing the logSomething directly inside of that child class.
I've tried this.logSomething(), I've tried adding it to a object, self executing (IIFE) function and everything I could thing of but to no result.
class Parent {
constructor() {}
logSomething() {
console.log('I am logging something')
}
}
class Child extends Paren {
logSomething() // This does not work
}
Currently doing this does not work, if throws a error referring to the fact that it things your trying to define a function.
I know it should be possible in some way, if I'm not mistaking React uses something similar with life-cycle methods right? Such as componentWillMount.
How would one go about doing this?
First error is that you are extending Paren instead of Parent.
Also you cannot just throw a random statement inside in a class. It needs to be inside a function.
If you want it to run whenever you create an instance of that class it should be inside the constructor or a function that gets called by it. (note that you need to call super() at the start of the constructor.
Finally, you still need to use this.logSomething or this.logSomething
class Parent {
constructor() {}
logSomething() {
console.log('I am logging something');
}
}
class Child extends Parent {
constructor() {
super();
this.logSomething(); // Will use Parent#logSomething since Child doesn't contain logSomething
super.logSomething(); // Will use Parent#logSomething
}
}
new Child();
class Parent {
constructor() {}
logSomething() {
console.log('Parent Logging Called');
}
}
class Child extends Parent {
constructor() {
super();
this.logSomething(); // Will call Child#logSomething
super.logSomething(); // Will call Parent#logSomething
}
logSomething() {
console.log('Child Logging Called');
}
}
new Child();
You could also do this:
class Parent {
constructor() {}
logSomething() {
console.log('Parent Logging Called');
}
}
class Child extends Parent {
logSomething() {
console.log('Child Logging Called and ...');
// Careful not use this.logSomething, unless if you are planning on making a recursive function
super.logSomething();
}
}
new Child().logSomething();
You can call any function or use any property of the parent class using this, as long as the new class doesn't have its own definition for that property.
Look here for more information.
class Parent {
constructor() {}
logSomething() {
console.log('I am logging something')
}
}
class Child extends Parent {
logSomething() {
super.logSomething(); // Call parent function
}
}
a) you can't call a function there, you can call a function within a function declared in a class
b) you need to use this.logSomething()
example:
class Parent {
constructor() {}
logSomething() {
console.log('I am logging something')
}
}
class Child extends Parent {
fn() {
this.logSomething() // This does work
}
}
new Child().fn()
See other answers for when fn is called logSomething in the child class - then you'd need super.logSomething() to call the "parent" logSomething instead of the child logSomething
I'm trying to write a subclass using the es6 class syntax. The subclass has some complicated logic to perform before calling the superclass constructor, so I tried factoring it out into a function. However, this seems to be impossible since this is not defined until after super() is called.
I've tried simply calling super() twice, once at the start of the constructor and once at the end, but it feels wrong and wastes the work that the superclass constructor does the first time.
class Parent {
constructor(x) {
console.log('some expensive thing with ' + x);
}
}
class Child extends Parent {
constructor() {
let x = this.f();
super(x);
}
f() {
// complicated logic
return 3;
}
}
let c = new Child();
Running the code as written results in ReferenceError: Must call super constructor in derived class before accessing 'this' or returning from derived constructor at new Child. Removing this and attempting to call f() results in ReferenceError: f is not defined at new Child.
Is there any way to factor the subclass constructor logic somewhere else, even if it's okay if this isn't bound?
I would use an initialization function separate from the constructor as that give you more control over when/if parent initialization happens.
class Parent {
constructor(x) {
this.init(x);
console.log("parent constructor does other stuff");
}
init(x) {
console.log("parent init runs")
}
}
class Child extends Parent {
constructor(x) {
super(x);
}
init(x) {
console.log("child init runs");
super.init(x); // This call is optional if you don't want to run the parent's init code
}
}
let c = new Child();
Using a static method could be a solution.
class Parent {
constructor(x) {
console.log('some expensive thing with ' + x);
}
}
class Child extends Parent {
constructor() {
let x = Child.f();
super(x);
}
static f() {
// complicated logic
return 3;
}
}
let c = new Child();
Is It good/bad practice to call a child method from a parent class?
class Parent {
constructor() {
// if 'autoPlay' exists (was implemented) in chain
if (this.autoPlay) {
this.autoPlay(); // execute from parent
}
}
}
class ChildA extends Parent {
autoPlay() {
console.log('Child');
}
}
class ChildB extends Parent {
// 'autoPlay' wasn't implemented
}
const childA = new ChildA();
const childB = new ChildB();
Is it a good practice to call a child method from a parent class?
Yes, it's a totally normal practise. The parent class just calls some method of the instance, and if the child class has overridden the method then the child method is called. However, you usually wouldn't do such a "has my instance defined this method" test, you just would call it. If you want to do nothing by default, just define an empty method (like in #scipper's answer). If you want to make the method abstract (force child classes to override it), you can either leave it undefined or define a method that throws an appropriate exception.
Is is a bad practice to call a child method from a parent constructor?
Yes. Don't do that. (It's a problem in all languages).
The purpose of a constructor is to initialise the instance and nothing else. Leave the invocations of side effects to the caller. This will ensure that all child constructors will finish their initialisation as well.
A contrived example:
class Parent {
autoPlay() {
this.play("automatically "); // call child method
}
play(x) {
console.log(x+"playing default from "+this.constructor.name);
}
}
class ChildA extends Parent {
// does not override play
}
class ChildB extends Parent {
constructor(song) {
super();
this.song = song;
}
play(x) {
console.log(x+"playing "+this.song+" from ChildB");
}
}
const child1 = new ChildA();
child1.autoPlay();
const child2 = new ChildB("'Yeah'");
child2.autoPlay();
Notice how that would not work if the Parent constructor did call autoplay. If you don't like to need an extra method call everywhere after the instantiation, use a helper function. It might even be a static method:
class Parent {
autoPlay() { … }
play { … }
static createAndAutoPlay(...args) {
const instance = new this(...args);
instance.autoPlay();
return instance;
}
}
…
const child1 = ChildA.createAndAutoPlay();
const child2 = ChildB.createAndAutoPlay("'Yeah'");
It would be better style to define an empty implementation of autoPlay in the Parent class, and override it in the child.
class Parent {
constructor() {
this.autoPlay();
}
autoPlay() {
}
}
class Child extends Parent {
autoPlay() {
console.log('Child');
}
}
const child = new Child();
class Parent {
print() {
console.log('hey i am parent');
}
}
class Child extends Parent {
constructor() {
super();
}
print() {
console.log('hey i am child');
}
}
x = new Parent();
console.log(Object.getPrototypeOf(x))
x.print();
Though the [[prototype]] of x is an empty object but still It can access the print() function which is defined in class Parent.
I cannot understand why Object.getPrototypeOf(x) is an empty object.
It's in there, just non-enumerable. Try this:
Object.getOwnPropertyNames(Object.getPrototypeOf(x));
// ["constructor", "print"]
class Parent {
print() {
console.log('hey i am parent');
}
}
class Child extends Parent {
constructor() {
super();
}
print() {
console.log('hey i am child');
}
}
x = new Parent();
console.log(Object.getOwnPropertyNames(Object.getPrototypeOf(x)));
x.print();
What makes you think it's empty? It does have constructor and print properties, they are however not enumerable and not displayed by default on the console. (And of course it does have a [[prototype]] link to Object.prototype, but how/whether that is displayed depends on your console as well).
To inspect them, have a look at Get functions (methods) of a class.