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ReferenceError: super() called twice in derived class constructor - JavaScript | MDN

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ReferenceError: Super constructor may only be called once (V8-based) ReferenceError: super() called twice in derived class constructor (Firefox) ReferenceError: 'super()' can't be called more than once in a constructor. (Safari)

Error type

ReferenceError

What went wrong?

The super() call can only be called at most once for each new call to a derived class constructor. This is because super() is responsible for initializing the parent class, and calling it more than once would result in the parent constructor being called multiple times.

The best way to prevent this is to ensure that super() is placed outside of any control flow structure. Otherwise, make sure that all code paths in the constructor lead to only one super() call.

The super() call can be "saved" in an arrow function nested within the constructor. Then, when you call the arrow function, you will also call super(), and the same rule applies: the arrow function can only be called at most once.

Examples

Invalid cases

js
class Base {}

class Derived extends Base {
  constructor() {
    super();
    super();
  }
}

Sometimes the bug may be more subtle.

js
class Base {
  constructor(flavor) {
    // Do something with the flavor
  }
}

class Derived extends Base {
  constructor(flavors) {
    if (flavors.includes("chocolate")) {
      super("chocolate");
    }
    if (flavors.includes("vanilla")) {
      super("vanilla");
    }
  }
}

Originally, flavors may never simultaneously include both "chocolate" and "vanilla", but if that ever happens, the constructor will call super() twice. You need to rethink about how your class should be structured to avoid this issue.

Valid cases

js
class Base {}

class Derived extends Base {
  constructor() {
    super();
    // More initialization logic
  }
}

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

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ReferenceError: super() called twice in derived class constructor

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