CODEWITHSUNDEEP
javascriptnode.jsclassoopecmascript-6
TSR
TSR

Reputation: 20406

How to call a super class method outside a class?

I have a class BMW that extends the class Car and override a method printModel:

class Car{
    constructor(model){
        this.model = model
    }
    printModel(){
        console.log(this.model)
    }
}

class BMW extends  Car{
    constructor(model, color){
        super(model);
        this.color = color
    }
    printModel(){
        console.log(this.model, this.color)
    }
}

let bmw = new BMW('F6', 'blue')
bmw.printModel() //will print : 'F6 blue'
bmw.super.printModel()// expected: 'F6' but not working

How to call a class super method on an instance of this BMW class?

Upvotes: 0

Views: 1968

Answers (5)

CertainPerformance
CertainPerformance

Reputation: 370689

You'll need a way to eventually get to super when calling something on the instantiation. One option is to define another method on BMW that calls super.printModel:

class Car {
  constructor(model) {
    this.model = model
  }
  printModel() {
    console.log(this.model)
  }
}

class BMW extends Car {
  constructor(model, color) {
    super(model);
    this.color = color
  }
  printModelBMW() {
    console.log(this.model, this.color)
  }
  printModelCar() {
    super.printModel();
  }
}

let bmw = new BMW('F6', 'blue')
bmw.printModelBMW() //will print : 'F6 blue'
bmw.printModelCar() // expected: 'F6'

Upvotes: 2

Estus Flask
Estus Flask

Reputation: 222369

If parent class method needs to be used alongside with child class method, this means class design went wrong, overriding parent method was a mistake.

It could be:

class Car{
    ...
    printGenericModel(){
        console.log(this.model)
    }
}

class BMW extends Car{
    ...
    printModel(){
        console.log(this.model, this.color)
    }
}

Or in case parent class can't be refactored:

class Car{
    ...
    printModel(){
        console.log(this.model)
    }
}

class BMW extends Car{
    ...
    get printGenericModel(){
        return super.printModel;
    }
    printModel(){
        console.log(this.model, this.color)
    }
}

Both methods are available as printGenericModel and printModel on BMW instance.

Upvotes: -1

Matt Browne
Matt Browne

Reputation: 12419

The super keyword can only be used inside a class, not outside. It's technically possible to call Car's printModel method directly, but this should really be avoided. The whole point of overriding a method in a superclass (like printModel) is to give it an implementation more appropriate to the subclass. So circumventing this is an indication of improper OO design.

However, for educational purposes I figure it would still be useful to know how this can technically be done, since it reveals how JS classes are really still using prototypes under the hood:

Car.prototype.printModel.call(bmw)

One way to improve the design to avoid this would be to add an optional argument to specify how you want to print the model:

class BMW extends Car{
    constructor(model, color){
        super(model)
        this.color = color
    }
    printModel(includeColor){
        if (includeColor) {
            console.log(this.model, this.color)
        }
        else super.printModel()
    }
}

Upvotes: 1

Osama
Osama

Reputation: 3040

class Car{
    constructor(model){
       this.model = model
    }
    printModel(){
    console.log(this.model)
    }
   getModel()
  {
      return this.model
  }

 }

  class BMW extends  Car{
      constructor(model, color){
        super(model);
         this.color = color
     }
     printModel(){
         console.log(super.getModel(), this.color)
     }
}

Upvotes: 0

Jake Holzinger
Jake Holzinger

Reputation: 6063

It's not possible to reference the super instance outside of the context of the class. If you really must use the method from the super instance from outside of the class, you can call the method yourself:

Car.prototype.printModel.call(bmw);

Upvotes: 2

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