I'm using ES6 classes and my class (A) extends class B and class B extends class C. How can A extend a method and then call C's version of that method.

class C {
  constructor() {
    console.log('class c');
  }
}

class B extends C {
  constructor() {
    super()
    console.log('no, I don\'t want this constructor.');
  }
}

class A extends B {
  constructor() {
    // What should I be doing here?  I want to call C's constructor.
    super.super();
  }
}

Edit: Thanks all, I'm going to stop trying to do this silly thing. The minor gains in code-re-use aren't worth the acrobatics in my situation.

I'm using ES6 classes and my class (A) extends class B and class B extends class C. How can A extend a method and then call C's version of that method.

class C {
  constructor() {
    console.log('class c');
  }
}

class B extends C {
  constructor() {
    super()
    console.log('no, I don\'t want this constructor.');
  }
}

class A extends B {
  constructor() {
    // What should I be doing here?  I want to call C's constructor.
    super.super();
  }
}

Edit: Thanks all, I'm going to stop trying to do this silly thing. The minor gains in code-re-use aren't worth the acrobatics in my situation.

• javascript
• ecmascript-6

Share Improve this question Follow edited Nov 25, 2024 at 21:47 Barmar 784k5757 gold badges548548 silver badges659659 bronze badges asked Mar 17, 2017 at 15:48 JulianJulian 2,85511 gold badge2323 silver badges3131 bronze badges 6

• 5 There is no code solution to this problem. You have an architecture problem, not a coding problem. – lonesomeday Commented Mar 17, 2017 at 15:56
• 1 You can't do that with classes. Please, explain your case in details, so the proper solution could be suggested. – Estus Flask Commented Mar 17, 2017 at 15:56
• 2 If B is a C, then B should be calling super in its constructor to be set up like C. That way A calling super gets both B and C's setups. But it sounds like B isn't the class you really want to extend from. If this isn't the case, then A should extend from C directly. – Joseph Commented Mar 17, 2017 at 15:58
• 3 Alternatively, just skip inheritance altogether in favor of position. – Joseph Commented Mar 17, 2017 at 16:02
• 1 you have just one problem. B has to call his super on his own constructor first, and A has to call his own super on his constructor first. so perhaps it's better working with an interface and not a master class? – mtizziani Commented Mar 17, 2017 at 16:05

 |  Show 1 more ment

5 Answers 5

Sorted by: Reset to default 5

You can’t not call the parent’s constructor in an ES6 class. Judging by your ment, maybe you should try something like this?

class Mixin {
  static include(constructor) {
    const descriptors = Object.getOwnPropertyDescriptors(Mixin.prototype);

    Object.keys(descriptors).forEach(name => {
      Object.defineProperty(constructor.prototype, name, descriptors[name]);
    });
  }

  someCommonFunction() {
    // Perform operation mon to B and C
  }
}

delete Mixin.prototype.constructor;

class B extends A {
}

Mixin.include(B);

class C extends A {
}

Mixin.include(C);

Simple but ugly way is to check for instance A in B's constructor:

class B extends C {
  constructor() {
    super()
    if (!(this instanceof A)) {
      console.log('no, no no, I don\'t want this!');  
    }
  }
}

class A {
  constructor() {
    console.log('class c');
  }
}

class B extends A {
  constructor() {
    super()
    if (!(this instanceof C)) {
      console.log('no, no no, I don\'t want this!');  
    }
  }
}

class C extends B {
  constructor() {
    super();
  }
}

const c = new C()

the best way i can think of, which might not necessarily be the best way, is the to define a function in B that calls A's super(), that way you will inherit that in C by extension and you will have it there ready to go.

class C {
  constructor(){
    console.log('C');
  }
}

class AB extends C {
  constructor(){
    super();
  }
}

class B extexts AB {
  constructor(){
    super();
    console.log('B');
  }
}

class A extends AB {
  constructor(){
    super();
  }
}

i think building a class between both and extend A and B from it is the only way because evere constructor has to call his own super constructor first.

new B(); // output is CB
new A(); // output is C

You need quite some acrobatics, and I don't think I would remend this form a 'clean / transparent code' point of view, however it can be done as follows:

<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">

<script type="text/javascript">

class C {
  constructor() {
      this.fooExtension = null;
  }

  initC(aFoo) {
    this.fooExtension = aFoo;
  }

  foo() {
    if (this.fooExtension === null) {
        console.log('fooFromC');
    } else {
        console.log('extend foo called from C');
        this.fooExtension();
    }
  }
}

class B extends C {
  constructor() {
    super();
  }

  foo() {
    super.foo();
  }
}

class A extends B {
  constructor() {
    super();

    let fooExtension = function() {
      console.log('fooFromA');
    };

    this.initC(fooExtension);
  }
}

c = new C();
c.foo();

a = new A();
a.foo();

</script>
</head>
<body>
</body>
</html>

This will result in the following output:

fooFromC
extend foo called from C
fooFromA