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SyntaxError: super() is only valid in derived class constructors - JavaScript | MDN

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Message

SyntaxError: 'super' keyword unexpected here (V8-based) SyntaxError: super() is only valid in derived class constructors (Firefox) SyntaxError: super is not valid in this context. (Safari)

Error type

SyntaxError

What went wrong?

The super() call is used to invoke the base constructor of a derived class, so the base class can initialize the this object. Using it anywhere else doesn't make sense.

super() can also be defined in an arrow function that's nested within the constructor. However, it cannot be defined in any other kind of function.

Examples

Invalid cases

You cannot call super() if the class has no extends, because there's no base class to call:

js
class Base {
  constructor() {
    super();
  }
}

You cannot call super() in a class method, even if that method is called from the constructor:

js
class Base {}

class Derived extends Base {
  constructor() {
    this.init();
  }

  init() {
    super();
  }
}

You cannot call super() in a function, even if the function is used as a constructor:

js
function Base(x) {
  this.x = x;
}

function Derived() {
  super(1);
}

Object.setPrototypeOf(Derived.prototype, Base.prototype);
Object.setPrototypeOf(Derived, Base);

Valid cases

You can call super() before calling any other method in the constructor:

js
class Base {}

class Derived extends Base {
  constructor() {
    super();
    this.init();
  }

  init() {
    // …
  }
}

You can call super() in an arrow function that's nested within the constructor:

js
class Base {}

class Derived extends Base {
  constructor() {
    const init = () => {
      super();
    };

    init();
  }
}

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This page was last modified on Jul 8, 2025 by MDN contributors.

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SyntaxError: super() is only valid in derived class constructors

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