Function.prototype[Symbol.hasInstance]()

value

The object to test. Primitive values always return false.

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 a instanceof test on value and the underlying target function.

TypeError

Thrown if this is not a bound function and this.prototype is not an object.

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.

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 a 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

• instanceof
• Symbol.hasInstance