[英]Ruby.Metaprogramming. class_eval
我的代碼中似乎有一個錯誤。 但是我只是找不到。
class Class
def attr_accessor_with_history(attr_name)
attr_name = attr_name.to_s
attr_reader attr_name
attr_writer attr_name
attr_reader attr_name + "_history"
class_eval %Q{
@#{attr_name}_history=[1,2,3]
}
end
end
class Foo
attr_accessor_with_history :bar
end
f = Foo.new
f.bar = 1
f.bar = 2
puts f.bar_history.to_s
我希望它返回一個數組[1,2,3]
。 但是,它不返回任何內容。
您不應該打開Class
添加新方法。 那就是模塊的用途。
module History
def attr_accessor_with_history(attr_name)
attr_name = attr_name.to_s
attr_accessor attr_name
class_eval %Q{
def #{attr_name}_history
[1, 2, 3]
end
}
end
end
class Foo
extend History
attr_accessor_with_history :bar
end
f = Foo.new
f.bar = 1
f.bar = 2
puts f.bar_history.inspect
# [1, 2, 3]
這是您可能要編寫的代碼(根據名稱判斷)。
module History
def attr_accessor_with_history(attr_name)
attr_name = attr_name.to_s
class_eval %Q{
def #{attr_name}
@#{attr_name}
end
def #{attr_name}= val
@#{attr_name}_history ||= []
@#{attr_name}_history << #{attr_name}
@#{attr_name} = val
end
def #{attr_name}_history
@#{attr_name}_history
end
}
end
end
class Foo
extend History
attr_accessor_with_history :bar
end
f = Foo.new
f.bar = 1
f.bar = 2
puts f.bar_history.inspect
# [nil, 1]
解:
class Class
def attr_accessor_with_history(attr_name)
ivar = "@#{attr_name}"
history_meth = "#{attr_name}_history"
history_ivar = "@#{history_meth}"
define_method(attr_name) { instance_variable_get ivar }
define_method "#{attr_name}=" do |value|
instance_variable_set ivar, value
instance_variable_set history_ivar, send(history_meth) << value
end
define_method history_meth do
value = instance_variable_get(history_ivar) || []
value.dup
end
end
end
測試:
describe 'Class#attr_accessor_with_history' do
let(:klass) { Class.new { attr_accessor_with_history :bar } }
let(:instance) { instance = klass.new }
it 'acs as attr_accessor' do
instance.bar.should be_nil
instance.bar = 1
instance.bar.should == 1
instance.bar = 2
instance.bar.should == 2
end
it 'remembers history of setting' do
instance.bar_history.should == []
instance.bar = 1
instance.bar_history.should == [1]
instance.bar = 2
instance.bar_history.should == [1, 2]
end
it 'is not affected by mutating the history array' do
instance.bar_history << 1
instance.bar_history.should == []
instance.bar = 1
instance.bar_history << 2
instance.bar_history.should == [1]
end
end
您將在Sergios Answer中找到解決問題的方法。 這里是一個解釋,您的代碼出了什么問題。
用
class_eval %Q{
@#{attr_name}_history=[1,2,3]
}
你執行
@bar_history = [1,2,3]
您在類級別而不是對象級別執行此操作。 變量@bar_history
在Foo對象中不可用,但在Foo類中不可用。
用
puts f.bar_history.to_s
您可以訪問-從不定義對象級別-屬性@bar_history。
在類級別定義閱讀器時,您可以訪問變量:
class << Foo
attr_reader :bar_history
end
p Foo.bar_history #-> [1, 2, 3]
@Sergio Tulentsev的答案行得通,但是它會提倡使用字符串評估的一種有問題的做法,當輸入值超出您的預期時,通常會帶來安全風險和其他意外情況。 例如,如果一個人打電話給Sergio,那會發生什么(不要嘗試):
attr_accessor_with_history %q{foo; end; system "rm -rf /"; def foo}
在沒有字符串評估的情況下,經常可以更仔細地進行ruby元編程。 在這種情況下,請使用簡單插值和閉包的define_method與instance_variable_ [get | set]一起發送:
module History
def attr_accessor_with_history(attr_name)
getter_sym = :"#{attr_name}"
setter_sym = :"#{attr_name}="
history_sym = :"#{attr_name}_history"
iv_sym = :"@#{attr_name}"
iv_hist = :"@#{attr_name}_history"
define_method getter_sym do
instance_variable_get(iv_sym)
end
define_method setter_sym do |val|
instance_variable_set( iv_hist, [] ) unless send(history_sym)
send(history_sym).send( :'<<', send(getter_sym) )
instance_variable_set( iv_sym, val @)
end
define_method history_sym do
instance_variable_get(iv_hist)
end
end
end
這是應該做的。 需要使用class_eval而不是在Class中定義attr_writer。
class Class
def attr_accessor_with_history(attr_name)
attr_name = attr_name.to_s
attr_reader attr_name
#attr_writer attr_name ## moved into class_eval
attr_reader attr_name + "_history"
class_eval %Q{
def #{attr_name}=(value)
@#{attr_name}_history=[1,2,3]
end
}
end
end
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