javascript - 对象的构造函数属性的奇怪行为

Question

在下面的代码中，为什么'scotty.constructor'设置为'Dog'函数？对象的构造函数属性一般是如何设置的？它是否设置为使用“new”创建对象的函数？如果是这样，则 scotty 是由“ShowDog”创建的，最后一行应该将其作为输出。

// Dog Constructor
function Dog(name, breed, weight) {
  this.name = name;
  this.breed = breed;
  this.weight = weight;
}


// Dog Prototype
Dog.prototype.species = 'Canine';

Dog.prototype.bark = function() {
  if (this.weight > 25) {
    console.log(this.name + ' says Woof!');
  } else {
    console.log(this.name + ' says Yip!');
  }
};

Dog.prototype.run = function() {
  console.log('Run!');
};

// ShowDog Constructor
function ShowDog(name, breed, weight, handler) {
  this.name = name;
  this.breed = breed;
  this.weight = weight;
  this.handler = handler;
}

// ShowDog Prototype
ShowDog.prototype = new Dog();

ShowDog.prototype.league = 'Webville';

ShowDog.prototype.stack = function() {
  console.log('Stack');
};

// ShowDog Instance
var scotty = new ShowDog('Scotty', 'Scotish Terrier', 15, 'Cookie');
console.log('Scotty\'s constructor is ' + scotty.constructor);

Answer 1

对象没有自己的 构造函数属性，它取自它们的protos。所以，如果没有这一行:

ShowDog.prototype = new Dog();

继承图是

一旦将new Dog分配给prototype(从而丢弃其当前值)，图片将变为:

并且构造函数值(Dog)取自scotty.__proto__.__proto__

要使继承按预期工作，您必须手动分配构造函数:

ShowDog.prototype = new Dog()
ShowDog.prototype.constructor = ShowDog

产生以下图表: