Question

我想在创建新对象时定义不同“类”/原型的多个属性。

class Animal
    constructor: (@name, @temperament, @diet) ->

    #methods that use those properties
    eat: (food) ->
        console.log "#{name} eats the #{food}."

class Bird extends Animal
    constructor: (@wingSpan) ->

    #methods relating only to birds

class Cat extends Animal
    constructor: (@tailLength) ->

    #methods relating only to cats

myCat = new Cat ("long", {"Mr. Whiskers", "Lazy", "Carnivore"})

但是，我做错了。只有Cat的构造函数似乎获得了任何属性。

另外，有没有办法用键/值对定义它们？理想情况下，我会编写类似myCat = new Cat (tailLength: "long", name: "Mr. Whiskers", temperament: "Lazy")的内容，以便我可以定义不按顺序排列的属性，如果我未能定义像“diet”这样的属性，它将回退到默认值。

我的理解是原型方法会冒出来，所以如果我调用myCat.eat 'cat food'，输出应该是"Mr. Whiskers eats the cat food."但是......它不可能是因为Animal类没有得到新猫的名字。

Answer 1

如果您的意思是“对象”，请使用{}。

class Animal
    constructor: ({@name, @temperament, @diet}) ->

    #methods that use those properties
    eat: (food) ->
        console.log "#{@name} eats the #{food}."

class Bird extends Animal
    constructor: ({@wingSpan}) ->
      super

    #methods relating only to birds

class Cat extends Animal
    constructor: ({@tailLength}) ->
      super

    #methods relating only to cats

myCat = new Cat(tailLength: "long", name: "Mr. Whiskers", temperament: "Lazy", diet: "Carnivore")

如何为多个CoffeeScript类定义属性

1 个答案: