Question

我正在学习如何在javascript中继承。我跟着this example。解释原型链。我创建了一个更详细的示例，显示了3个级别Person > Employee > Engineer 我的例子的不同之处在于创建了Employee的实例，传递了参数。但是，如果我按照上面提到的mozilla示例，我需要将其传递给Child原型：

 var Engineer.prototype = new(Employee);

我没有传递任何论据。

之后，当我将Engineer.prototype内容打印到控制台时，我确实看到了该物业＆＃34;公司＆＃34; as＆＃34; undefined＆＃34;。

Engineer prototype { _alive: true,
  _company: undefined,
  _title: null,
  constructor: [Function] }
Drew instance:  { _alive: true,
  _company: 'Apple',
  _title: null,
  _specialization: 'tester' }

我确定可以忍受这个，但它看起来并不干净整洁。

这里的问题是什么是最佳做法？ Shell我宁愿不将任何参数传递给构造函数并创建一些init方法并显式调用它或shell我忽略构造函数中的未定义属性？

这是我的例子中的代码：

var Person = function(){
  this._alive = true;
};


var Employee = function(company){
  Person.call(this);
  this._company = company;
  this._title = null;
}

Person.prototype.die = function(){
  this._alive = false;
}


Employee.prototype = new Person();
Employee.prototype.constructor = Employee;

Employee.prototype.setTitle = function(newtitle){
  this._title = newtitle;
}
Employee.prototype.title = function(){
  return this._title;
}
Employee.prototype.company = function(){
  return this._company;
}

var Engineer = function(company, specialization){
  Employee.call(this,company);
  this._specialization = specialization;
}

Engineer.prototype = new(Employee);
Engineer.prototype.constructor = Engineer;

var drew = new(Engineer)('Apple','tester');

console.log('Engineer prototype', Engineer.prototype);    
console.log('Drew instance: ',drew);

javascript继承挑战

0 个答案: