为什么将prototype的构造函数设置为其构造函数？

Question

关于这个脚本的一行:

function Vehicle(hasEngine, hasWheels) {
    this.hasEngine = hasEngine || false;
    this.hasWheels = hasWheels || false;
}

function Car (make, model, hp) {
    this.hp = hp;
    this.make = make;
    this.model = model;
}

Car.prototype = new Vehicle(true, true);
Car.prototype.constructor = Car; 
Car.prototype.displaySpecs = function () {
    console.log(this.make + ", " + this.model + ", " + this.hp + ", " + this.hasEngine + ", " + this.hasWheels);
}

var myAudi = new Car ("Audi", "A4", 150);
myAudi.displaySpecs(); // logs: Audi, A4, 150, true, true

我的问题是:做什么

Car.prototype.constructor = Car;  

做？更重要的是,不这样做的后果是什么,在哪种情况下最有用？

Answer 1

它恢复了.constructor您覆盖的原始原型对象上的属性.人们恢复它,因为它预计会在那里.

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if (my_obj.constructor === Car) { ... }

这不是必需的,因为它instanceof是一个更好的测试IMO.

if (my_obj instanceof Car) { ... }

if (my_obj instanceof Vehicle) { ... }