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Message

SyntaxError: 'super' keyword unexpected here (V8-based)
SyntaxError: use of super property accesses only valid within methods or eval code within methods (Firefox)
SyntaxError: super is not valid in this context. (Safari)

The super.x syntax is used to access properties on the prototype of the current object. It can be used in methods of both object literals and classes, field initializers, and static initialization blocks, but not in other contexts.

Examples

Invalid cases

You can't use super.x outside of a method in an object:

const obj = {
  __proto__: { x: 1 },
  x: super.x, // SyntaxError: use of super property accesses only valid within methods or eval code within methods
};

You can't use super.x in a function, even if that function has the effect of being a method:

function getX() {
  return super.x; // SyntaxError: use of super property accesses only valid within methods or eval code within methods
}

const obj = {
  getX,
  getX2: function () {
    return super.x; // SyntaxError: use of super property accesses only valid within methods or eval code within methods
  },
};

class Derived extends Base {
  getX = () => super.x;
}

Valid cases

You can use super.x in a method:

class Base {
  x = 1;
}

class Derived extends Base {
  getX() {
    return super.x;
  }
}

You can use super.x in a field initializer:

class Derived extends Base {
  x = super.x;
}

You can use super.x in object methods too:

const obj = {
  __proto__: { x: 1 },
  getX() {
    return super.x;
  },
};

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

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SyntaxError: use of super property/member accesses only valid within methods or eval code within methods

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Message

Error type

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Examples

Invalid cases

Valid cases

See also

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