Function.prototype[Symbol.hasInstance]()

¥Syntax

func[Symbol.hasInstance](value)

¥Parameters

value

要测试的对象。原始值始终返回 false。

¥Return value

如果 func.prototype 在 value 的原型链中，则 true；否则，false。如果 value 不是对象或 this 不是函数，则始终返回 false。如果 this 是 绑定函数，则返回对 value 和底层目标函数进行 instanceof 测试的结果。

¥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

如果 this 不是绑定函数并且 this.prototype 不是对象，则抛出该异常。

¥Description

只要存在这样的方法，instanceof 运算符就会调用右侧的 [Symbol.hasInstance]() 方法。因为默认情况下所有函数都继承自 Function.prototype，所以它们都会有 [Symbol.hasInstance]() 方法，所以大多数时候，Function.prototype[Symbol.hasInstance]() 方法指定当右侧是函数时 instanceof 的行为。该方法实现了 instanceof 操作符的默认行为（当 constructor 没有 [Symbol.hasInstance]() 方法时，算法相同）。

¥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).

与大多数方法不同，Function.prototype[Symbol.hasInstance]() 属性不可配置且不可写。这是一项安全功能，可防止绑定函数的底层目标函数不可获取。请参阅 这个 StackOverflow 答案 的示例。

¥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

你很少需要直接调用此方法。相反，该方法由 instanceof 运算符调用。你应该期望这两个结果通常是相等的。

¥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

如果你想调用默认的 instanceof 行为，但你不知道构造函数是否具有重写的 [Symbol.hasInstance]() 方法，你可能需要使用此方法。

¥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

¥See also

• instanceof
• Symbol.hasInstance