If I create the following classes, is there any way in Class1 to detect when the instance is actually one of Class2, without knowing anything about Class2?
i.e. Can Class1 tell when it's the parent class being extended?
class Class1 {
constructor() {
// Code to detect whether parent here
}
}
class Class2 extends Class1 {
constructor() {
super();
}
}
This is what new.target was made for - it gives you the constructor that new was invoked with. So
class Class1 {
constructor() {
if (new.target != Class1) {
// Class1 is used as a parent class
}
}
}
Related
I am making an application using JavaScript. One thing that I noticed is that when I call super() from child class constructor, I cannot use child class fields.
For example:
class Parent {
constructor() {
console.log(this.x);
}
}
class Child extends Parent {
x = 5;
constructor() {
super();
}
}
Output: undefined
As you see, the parent constructor cannot access the x field of the child constructor. There are similar questions on stackoverflow like:
Access JavaScript class property in parent class
The solution they give is to use an init() method on the Parent class. I realized that every method other than the constructor of the Parent class has access to child fields.
class Parent {
init() {
console.log(this.x);
}
}
class Child extends Parent {
x = 5;
constructor() {
super();
this.init();
}
}
Output: 5
However, I can also get access to the x field by using Timeout with delay of 0:
class Parent {
constructor() {
setTimout(() => console.log(this.x), 0);
}
}
class Child extends Parent {
x = 5;
constructor() {
super();
}
}
Output: 5
All the answers on stackoverflow give the solution, but nobody explains how and why this happens. So my question is:
1. Why the constructor doesn't have access to fields of the child class?
2. Why adding a timeout function gives access to the fields of the child class?
So, I'm new to Typescript, and recently started learning from documentation. I was looking at its documentation and there were no signs of reusing method from other class.
Class A file
export class A{
... constuctor(){
const queue = this.createQueue()
}
createQueue(){
console.log("Has access to this", +this);
}
}
And there is a class B with exact same definations and uses the same method. How do I make this reusable so that, both of them call with "this"?
One solution I thought of is to create a seperate helper Class that could I use, but I'm not sure how to do that.
Any thoughts?
In this case, inheritance would probably be your best bet. This basically means that you can extend upon another class and include all of the other one's properties/getters/setters/methods.
// we denote this class as abstract because it must be extended and cannot be directly initialized, i.e. new BaseFoo()
abstract class BaseFoo<T> { // `T` is the item type within the queue
queue: T[];
constructor() {
this.queue = this.createQueue();
}
createQueue() {
// can access `this`
return [];
}
}
class FooA extends BaseFoo<number> {
constructor() {
super(); // runs child class's constructor
console.log(this.queue); // []
}
}
class FooB extends BaseFoo<string> {
constructor() {
super();
console.log(this.queue); // []
}
}
TypeScript Playground Link
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
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();
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();