Question

我很困惑为什么在更改对象的构造函数属性以指向另一个函数之后，对象仍然是旧构造函数的实例

//Original Constructor
      function orig_cons() {}

//New Constructor
  function new_cons(){}

 //Adding property to myfun prototype
  new_cons.prototype.x = 1;

 //Invoking a new object of 1st constructor
  var obj1 = new orig_cons();

//check obj1 instanceof 1st constructor
 console.log(obj1 instanceof orig_cons); // true

 //Changing constructor property to point to 2nd function
  obj1.constructor = new_cons;

//check obj1 instanceof 2nd function
 console.log(obj1 instanceof new_cons); // false

obj1的prototype属性仍然是原始构造函数的原型：

       console.log(Object.getPrototypeOf(obj1)); //orig_cons {}

Answer 1

instanceof检查对象的原型而不是构造函数，举个例子：

function Person(name){
  this.name = name;
}

var  p = new Person();

console.log(p instanceof Person); //true

function Func(){}

p.constructor = Func;

console.log(p instanceof Person); // true

Person.prototype = {};

console.log(p instanceof Person); // false

注意更改原型如何影响instanceof。

来自MDN：https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/instanceof

instanceof运算符测试了constructor.prototype的存在对象的原型链。

在更改构造函数属性之后，Object仍然是原始构造函数的实例

1 个答案: