Function.prototype[Symbol.hasInstance]()

The [Symbol.hasInstance]() method of {{jsxref("Function")}} instances specifies the default procedure for determining if a constructor function recognizes an object as one of the constructor's instances. It is called by the instanceof operator.

Syntax#

func[Symbol.hasInstance](value)

Parameters#

• value
• : The object to test. Primitive values always return false.

Return value#

true if func.prototype is in the prototype chain of value; otherwise, false. Always returns false if value is not an object or this is not a function. If this is a bound function, returns the result of an instanceof test on value and the underlying target function.

Exceptions#

• {{jsxref("TypeError")}}
• : Thrown if this is not a bound function and this.prototype is not an object.

Description#

The instanceof operator calls the [Symbol.hasInstance]() method of the right-hand side whenever such a method exists. Because all functions inherit from Function.prototype by default, they would all have the [Symbol.hasInstance]() method, so most of the time, the Function.prototype[Symbol.hasInstance]() method specifies the behavior of instanceof when the right-hand side is a function. This method implements the default behavior of the instanceof operator (the same algorithm when constructor has no [Symbol.hasInstance]() method).

Unlike most methods, the Function.prototype[Symbol.hasInstance]() property is non-configurable and non-writable. This is a security feature to prevent the underlying target function of a bound function from being obtainable. See this Stack Overflow answer for an example.

Examples#

Reverting to default instanceof behavior#

You would rarely need to call this method directly. Instead, this method is called by the instanceof operator. You should expect the two results to usually be equivalent.

class Foo {}
const foo = new Foo();
console.log(foo instanceof Foo === Foo[Symbol.hasInstance](foo)); // true

You may want to use this method if you want to invoke the default instanceof behavior, but you don't know if a constructor has an overridden [Symbol.hasInstance]() method.

class Foo {
  static [Symbol.hasInstance](value) {
    // A custom implementation
    return false;
  }
}

const foo = new Foo();
console.log(foo instanceof Foo); // false
console.log(Function.prototype[Symbol.hasInstance].call(Foo, foo)); // true

Specifications#

{{Specifications}}

Browser compatibility#

{{Compat}}

See also#

• instanceof
• {{jsxref("Symbol.hasInstance")}}