Question

我正在尝试一种将动态原型模式与原型链相结合的继承模型。这项调查纯粹是学术性的，试图理解JavaScript中对象的属性。我相信可以推荐更好的模型。我很好奇为什么这两种方法的组合无法生成我正在寻找的原型链。

function Animal(params) {
  this.species = undefined;
  this.sound   = undefined;
  if (params.species !== undefined){ this.species = params.species; }
  if (params.sound   !== undefined){ this.sound   = params.sound;   }
  if (Animal.prototype.makeSound === undefined) {
    Animal.prototype.makeSound = function() {
      console.log(this.sound);
    };
  }
}

function Bear(params) {
  if (params.species === undefined){ params.species = "Bear"; }
  if (params.sound   === undefined){ params.sound   = "Grr";  }
  Animal.call(this, params);
  if (Bear.prototype.swipe === undefined) {
    Bear.prototype = new Animal(params);
    Bear.prototype.swipe = function() {
      console.log("swipe - '" + this.sound + "'");
    };
  }
}
var bear   = new Bear({});
bear.swipe();     // Uncaught TypeError: bear.swipe is not a function(…)
bear.makeSound(); // Uncaught TypeError: bear.makeSound is not a function(…)

如果我将原型定义移到构造函数之外（即不再是动态模式），这可以正常工作 - 删除链接调用中的参数＆amp;检查动物构造函数中的params使用：

function Animal(params) {
  this.species = undefined;
  this.sound   = undefined;
  if (params !== undefined) {
    if (params.species !== undefined){ this.species = params.species; }
    if (params.sound   !== undefined){ this.sound   = params.sound;   }
  }
}
Animal.prototype.makeSound = function() {
  console.log(this.sound);
}

function Bear(params) {
  if (params.species === undefined){ params.species = "Bear"; }
  if (params.sound   === undefined){ params.sound   = "Grr";  }
  Animal.call(this, params);
}
Bear.prototype = new Animal();
Bear.prototype.swipe = function() {
  console.log("swipe - '" + this.sound + "'");
}
var bear   = new Bear({});
bear.swipe();     // swipe - 'Grr'
bear.makeSound(); // Grr

我认为这是不打扰实际项目中的动态模式的原因。我只是好奇。

我想要做的完整示例是here

无法将动态原型模式与原型链接相结合

0 个答案: