JavaScript - 为什么设置原型(prototype)原型(prototype)不起作用？

Question

据我所知，JavaScript 中的继承可以通过以下方式完成(复制自 MDN ):

// Shape - superclass
function Shape() {
  this.x = 0;
  this.y = 0;
}

// superclass method
Shape.prototype.move = function(x, y) {
  this.x += x;
  this.y += y;
  console.info('Shape moved.');
};

// Rectangle - subclass
function Rectangle() {
  Shape.call(this); // call super constructor.
}

// subclass extends superclass
Rectangle.prototype = Object.create(Shape.prototype);
Rectangle.prototype.constructor = Rectangle;

var rect = new Rectangle();

console.log('Is rect an instance of Rectangle? ' + (rect instanceof Rectangle)); // true
console.log('Is rect an instance of Shape? ' + (rect instanceof Shape)); // true
rect.move(1, 1); // Outputs, 'Shape moved.'

我不明白的是为什么要替换:

Rectangle.prototype = Object.create(Shape.prototype);
Rectangle.prototype.constructor = Rectangle;

与:

Rectangle.prototype.prototype = Shape.prototype;

没有完成同样的事情吗？

对实例执行类似的操作似乎效果很好。示例:

var rect = new Rectangle();
rect.__proto__.__proto__ = Shape.prototype;

但是以这种方式操作原型(prototype)是 discouraged

Answer 1

因为inst.__proto__ == Rectangle.prototype。如果您想操作 .prototype 对象的原型(prototype)(即它继承的内容)，则需要使用

Rectangle.prototype.__proto__ = Shape.prototype;