以下内容来自The Rspec Book.我很奇怪,为什么messages << message在def puts不是@messages << message.我知道这||= []是memoization,我也理解这@messages是一个实例变量.那么,它应该在同一个类中使用吗?或者这是一个错字?
class Output
def messages
@messages ||= []
end
def puts(message)
messages << message
end
end
def output
@output ||= Output.new
end
Given /^I am not yet playing$/ do
end
When /^I start a new game$/ do
game = Codebreaker::Game.new(output)
game.start
end
Then /^I should see "([^"]*)"$/ do |message|
output.messages.should include(message)
end
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因为
def puts(message)
messages << message
end
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然后打电话
def messages
@messages ||= []
end
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的||= []受让人空数组到@messages如果不存在它.这就是为什么puts(message)不@messages直接使用.
编辑:如果你@messages << message直接做@messages,那时可能还不存在.这就是你经历的原因messages.
我认为你的混淆在@messages变量和messages方法之间
class Output
def messages
@messages ||= []
end
def puts(message)
messages << message #=> here `messages` refer to the method above
end
end
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这是一个简单的例子
class Output
def m
@messages ||= []
end
def puts(message)
m << message #=> method `m`
end
end
v = Output.new
v.puts "blah"
v.m #=> ["blah"]
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示例2(为了更加清晰)
class Output
def m
@messages ||= []
end
def puts(message)
m << message #=> here `m` refer to the method above
end
def show_at_messages_variable
@messages
end
end
v = Output.new
v.puts "blah"
v.show_at_messages_variable #=> ["blah"]
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作者选择了惰性对象创建,消息不是在构造函数中创建的,而是在getter方法中创建的.只要您不直接访问@messages,就无法获得零例外.
class Output
def messages
@messages ||= []
end
def puts(message)
# If i was to use the variable @messages at this point,
# i would get a nil exception, the array was never allocated.
# Using the getter ensures that the array is allocated.
messages << message
end
end
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如果你有一个明显的不需要懒惰创建对象,你可以用以下方式编写同一个类.
class Output
attr_reader :messages
def initialize(message)
@messages = []
end
def puts(message)
# messages was created when this object was initialized,
# thus it is perfectly safe to use here.
@messages << message
end
end
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