Function.prototype[Symbol.hasInstance]()
Baseline Widely available
This feature is well established and works across many devices and browser versions. It’s been available across browsers since July 2015.
The [Symbol.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.
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 a instanceof
test on value
and the underlying target function.
Exceptions
TypeError
-
Thrown if
this
is not a bound function andthis.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 StackOverflow 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 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
Specifications
Specification |
---|
ECMAScript Language Specification # sec-function.prototype-%symbol.hasinstance% |
Browser compatibility
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