JavaScript 在子类中设置构造函数？

Question

下面是一个简单的 JavaScript 继承示例。该线路的目的是什么 从下面？

Dog.prototype.constructor = Dog;

如果我取消注释或注释掉，代码似乎会给出相同的输出。

function Animal(age){
    this.age = age;
    console.log("Animal constructor called")
}
Animal.prototype.walk = function(){
     console.log("Animal with age " + this.age + " walked"); 
}

function Dog(color, age){
    this.color = color;
    Animal.call(this, age)
    console.log("Dog constructor called")
}

Dog.prototype = new Animal();      
//Dog.prototype.constructor = Dog;

dog = new Dog("white", 10);
console.log("Age: " + dog.age +  " Color: " + dog.color );

Answer 1

定义函数时，会自动创建 prototype 属性:

function Animal(age){
    this.age = age;
    console.log("Animal constructor called")
}

Animal.prototype.walk = function(){
     console.log("Animal with age " + this.age + " walked"); 
}

了解了构造函数，引用原型(prototype)对象就很容易了:Animal.prototype。

从prototype对象端来看，它有一个constructor属性来访问原始函数:Animal.prototype.constructor === Animal 。这对于从实例确定它的原始类是什么很有用:

var lion = new Animal();
console.log(lion.constructor === Animal) // prints "true"
console.log(lion.__proto__.constructor === Animal) // prints "true"

<小时/>

如果是继承，则需要手动设置构造函数:

Dog.prototype = new Animal();     
console.log(Dog.prototype.constructor) // prints "function Animal(){}"
Dog.prototype.constructor = Dog;       // setup manually the constructor to Dog
console.log(Dog.prototype.constructor) // prints "function Dog(){}"

为了稍后验证它的正确性:

var dog = new Dog("white", 10);

console.log(dog.constructor === Animal)           // prints "false"
console.log(dog.__proto__.constructor === Animal) // prints "false"
console.log(dog.constructor === Dog)              // prints "true"
console.log(dog.__proto__.constructor === Dog)    // prints "true"

<小时/>

简而言之，如果您想使用构造函数来验证哪个类具有特定实例，则应该修复构造函数。