Question

为什么在扩展它时需要在.prototype中使用Shape.prototype？

// Shape — superClass
function Shape() {
  this.x = 0;
  this.y = 0;
  }

Shape.prototype.move = function(x, y) {
  this.x += x;
  this.y += y;
  console.info('Figure has rode out somewhere.');
};

function Rectangle() {
  Shape.call(this); 
}

Rectangle.prototype = Object.create(Shape.prototype);//<<<=HERE

我的意思是这个＆＃39; Shape.prototype＆＃39;。为什么我需要使用原型而不仅仅是Shape？据我所知.prototype包含一个类的继承属性和方法。我的Shape类是基本类，没有继承属性。

Answer 1

通常会将方法添加到对象的基础原型中，以便不必将它们存储在稍后生成的每个实例中。在此示例中就是这种情况，move()函数存储在Shape.prototype中。

如果您未将Shape.prototype与Object.create()一起使用，则不会将附加到该对象的方法继承到Rectangle。

从技术上讲，你可以只使用Object.create(Shape)，但仅限于Shape只有实例属性直接附加到它的情况下，即使在这种情况下，它也不是一个非常可扩展的解决方案，因为如果你曾经决定在将来返回并向Shape.prototype添加方法，没有任何子类型会继承它。

// Shape — superClass
function Shape() {
  this.x = 0;
  this.y = 0;
}

// Shape.prototype is a unique and specific instance of Object
// This particular instance is what Shape inherits from and is
// the reason why Shape will get a .toString() and a .valueOf()
// method.
console.log(Shape.prototype);

// This method is added to Shape.prototype. If you don't inherit
// from this, you won't inherit this method!!
Shape.prototype.move = function(x, y) {
  this.x += x;
  this.y += y;
  console.info('Figure has rode out somewhere.');
};

function Rectangle() {
  Shape.call(this); 
}

// If we don't inherit from Shape.prototype, we won't get
// any of the methods attached to it!
Rectangle.prototype = Object.create(Shape); 

var r = new Rectangle();

console.log(r.x, r.y);  // Works just fine for instance properties attached direclty to Shape
r.move(10,20);          // Fails because Shape.prototype wasn't used for inheritance!

扩展时为什么我需要在superClass.prototype中使用.prototype？

1 个答案: