Question

我刚刚开始学习CoffeeScript，我想知道从子实例中检索类中的静态属性的最佳实践。

class Mutant
    MutantArray: []

    constructor: (@name, @strength = 1, @agility = 1) ->
        @MutantArray.push(@name)

    attack: (opponent) ->
        if opponent in @MutantArray then console.log @name + " is attacking " + opponent else console.log "No Mutant by the name of '" + opponent + "' found."


    @getMutants: () ->
        # IS THIS RIGHT?
        console.log @.prototype.MutantArray

Wolverine = new Mutant("Wolverine", 1, 2)
Rogue = new Mutant("Rogue", 5, 6)

Rogue.attack("Wolverine")

Mutant.getMutants()

我希望我的getMutants（）方法是静态的（不需要实例化）并返回已实例化的Mutant名称列表。 @ .prototype.MutantArray似乎工作正常，但是有更好的方法吗？我试过@MutantArray，但这不起作用。

谢谢！

Answer 1

我认为您应该将MutantArray定义为静态字段。然后，从非静态方法中，您应该通过类和静态方法引用它，通过@访问它。像这样：

class Mutant
    @MutantArray: []

    constructor: (@name, @strength = 1, @agility = 1) ->
         Mutant.MutantArray.push(@name)

    attack: (opponent) ->
        if opponent in Mutant.MutantArray then console.log @name + " is attacking " + opponent else console.log "No Mutant by the name of '" + opponent + "' found."


    @getMutants: () ->
        # IS THIS RIGHT?
        console.log @MutantArray

Answer 2

我认为是这样的：

class Mutant
    MutantArray: []

    constructor: (@name, @strength = 1, @agility = 1) ->
        @MutantArray.push(@name)

    attack: (opponent) ->
        if opponent in @MutantArray then console.log @name + " is attacking " + opponent else console.log "No Mutant by the name of '" + opponent + "' found."


    getMutants: () ->
        # IS THIS RIGHT?
        console.log @.MutantArray

Wolverine = new Mutant("Wolverine", 1, 2)
Rogue = new Mutant("Rogue", 5, 6)

Rogue.attack("Wolverine")

Mutant.getMutants()

getMutants必须是原型方法，并使用@ .getMutants

检索数组值

CoffeeScript - 如何在类中检索静态数组属性

2 个答案: