javascript - 实例化对象中不存在作为属性的子类方法

Question

我正在编写一个状态模式，而作为子类编写的不同状态的属性的方法似乎不存在于实例化的 State 对象中。

帮忙？

代码中不起作用的部分是这样的:

var Main,
    Active,
    Inactive;

Main = (function() {
  function Main() {
    // Construct Main object
    this.currentStatus = new Inactive();
  }
  return Main;

})();

Active = (function() {
  function Active() {
    // Construct Active object
  }
  Active.prototype.deactivate = function() {
    // Deactivate
  }
  Active.prototype.activate = function() {
    // Do nothing
  }
  return Active;
})();
Active.prototype = Object.create(Main.prototype);
Active.prototype.constructor = Active;

Inactive = (function() {
  function Inactive() {
    // Construct Inactive object
  }
  Inactive.prototype.deactivate = function() {
    // Do nothing
  }
  Inactive.prototype.activate = function() {
    // Activate
  }
  return Inactive;
})();
Inactive.prototype = Object.create(Main.prototype);
Inactive.prototype.constructor = Inactive;

var object = new Main();

// This doesn't work
object.currentStatus.activate;

Answer 1

object.currentStatus 中没有 activate 方法，因为您要在第 41 行重新分配 Inactive.prototype。您需要执行此操作< em>在使用更多方法扩展原型(prototype)之前:

Inactive = (function() {
  function Inactive() {
    // Construct Inactive object
  }

  Inactive.prototype = Object.create(Main.prototype);
  Inactive.prototype.constructor = Inactive;

  Inactive.prototype.deactivate = function() {
    // Do nothing
  }
  Inactive.prototype.activate = function() {
    // Activate
  }
  return Inactive;
})();

演示:http://jsbin.com/beqofigihu/edit?js,console

Active 类也是如此。