The [@@hasInstance]()
method of 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.
The [@@hasInstance]()
method of 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.
func[Symbol.hasInstance](value)
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 [@@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 [@@hasInstance]()
method, so most of the time, the Function.prototype[@@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 @@hasInstance
method).
Unlike most methods, the Function.prototype[@@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 StackOverflow 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 [@@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
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@@hasInstance |
50 | 15 | 50 | 37 | 10 | 50 | 50 | 37 | 10 | 5.0 | 50 | 1.0 | 6.5.0 |
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https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/@@hasInstance