javascript - 如何正确设置原型(prototype)？

Question

我正在练习一本名为 Object-Oriented JavaScript 的书中的练习。 .

第 5 章的练习包含以下 4 个问题:

1. Create an object called shape that has the type property and a getType() method.
2. Define a Triangle() constructor function whose prototype is shape. Objects created with Triangle() should have three own properties—a, b, and c, representing the lengths of the sides of a triangle.
3. Add a new method to the prototype called getPerimeter().
4. Test your implementation with the following code:

.

  var t = new Triangle(1, 2, 3);
  t.constructor === Triangle;  // true
  shape.isPrototypeOf(t);      // true
  t.getPerimeter();            // 6
  t.getType();                 // "triangle"

这是我对上述问题的解决方案:

var shape = {
    type: 'triangle',
    getType: function() {
        return this.type;
    }
};

function Triangle(a, b, c) {
    this.a = a;
    this.b = b;
    this.c = c;
}

Triangle.prototype = shape;  // Maybe something goes wrong here

Triangle.prototype.getPerimeter = function() {
    return this.a + this.b + this.c;
}

// ======================================================
var t = new Triangle(1, 2, 3);              
console.log(t.constructor === Triangle);  // false
console.log(shape.isPrototypeOf(t));      // true
console.log(t.getPerimeter());            // 6
console.log(t.getType());                 // triangle

为什么console.log(t.constructor === Triangle);输出false作为结果？

我尝试删除 console.log(t.constructor === Triangle);，这使得 t.constructor === Triangle 等于 true，但会导致 TypeError: t.getType is not a function 错误。

怎样才能让我的执行结果和本书提供的答案一样？

Answer 1

添加

Triangle.prototype.constructor = Triangle;

之后

Triangle.prototype = shape;

解决了我的问题。

在第 6 章第 173 页，书上说

Overwriting the prototype has side effects on the constructor property. Therefore, it's a good idea to reset the constructor after inheriting