。每个元素都不能正确迭代数组元素

我有两个ruby类用于在游戏中存储一些自定义日志。 RoundLog包含与单轮相关的信息，而BattleLog只是包含多个RoundLog元素的数组。。每个元素都不能正确迭代数组元素

class RoundLog 
    attr_writer :actions, :number 

    def initialize(number) 
    @number = number 
    @actions = [] 
    end 
    def to_a 
    [@number, @actions] 
    end 
    ... 
end 

class BattleLog 
    attr_accessor :rounds 
    def initialize 
    @rounds = [] 
    end 
    def print 
    @rounds.each do |round| 
     round.to_a 
    end 
    end 
    ... 
end

如果我有以下BattleLog实例：

report = [#<RoundLog:0x00000008ab8328 @number=1, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>, 
#<RoundLog:0x00000008acc170 @number=2, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>, 
#<RoundLog:0x00000008aef5f8 @number=3, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>, 
#<RoundLog:0x00000008b02978 @number=4, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>, 
#<RoundLog:0x00000008b1a280 @number=5, @actions=["Rat hits Test and deals 1 points of damage"]>]

然后将下面这段代码是不工作：report.each {|x| x.to_a} 而是像返回格式正确的信息：

[1, ["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"], 
[2, ["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"], ...]

它返回整个RoundLog对象：

[#<RoundLog:0x00000008ab8328 @number=1, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>, 
#<RoundLog:0x00000008acc170 @number=2, @actions=["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"]>,...]

但是，如果我尝试这样的：report.first.to_a它正确返回[1, ["Rat hits Test and deals 1 points of damage", "Test hits Rat and deals 1 points of damage"] 任何想法我的代码有什么问题吗？我尝试将to_a重命名为其他内容，所以我不认为问题出现在函数名称中。这是我的第一个问题，所以请放纵。

来源

2014-12-25 switowski

使用map而不是each应该可以解决您的问题。

each在块内部运行一些操作，然后返回对象/ array/hash/enumerable /无论each被调用。然而，map会返回一个新数组，其中返回的值是在您的块中计算的。

来源

2014-12-25 22:00:05 Humza

谢谢，它的工作:)我真的错过了'each'和'map'之间的重要区别（我认为问题在别的地方）。 – switowski

。每个元素都不能正确迭代数组元素

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